DAMS  AND  WEIES 


AN  ANALYTICAL  AND  PRACTICAL  TREATISE  ON  GRAVITY 

DAMS  AND  WEIRS;  ARCH  AND  BUTTRESS  DAMS; 

SUBMERGED  WEIRS;  AND  BARRAGES 


By  W.  G.  BLIGH 

INSPECTING    ENGINEER    OP   IRRIGATION    WORKS 

DEPARTMENT   OF   INTERIOR,   CANADA 
MEMBER,  INSTITUTE  CIVIL   ENGINEERING    (LONDON) 


ILLUSTRATED 


AMERICAN  TECHNICAL  SOCIETY 

CHICAGO 

1915 


COPYRIGHT,  1916,  BY 

AMERICAN  TECHNICAL  SOCIETY 



COPYRIGHTED  IN  GREAT  BRITAIN 
ALL  RIGHTS  RESERVED 


CONTENTS 


Gravity  dams 2 

Pressure  of  water  on  wall 2 

Method  for  graphical  calculations 2 

Conditions  of  '  'middle  third' '  and  limiting  stress 3 

Compressive  stress  limit 4 

Design  of  dams 4 

Theoretical  profile 4 

Practical  profile 8 

Crest  width 9 

Rear  widening 10 

Variation  of  height 13 

High  and  wide  crest 13 

Graphical  method .- .  ^ 16 

Analytical  method if. 18 

Pressure  distribution 23 

Graphical  method  for  distribution  of  pressure 25 

Maximum  pressure  limit 27 

Limiting  height 29 

Internal  shear  and  tension 30 

Security  against  failure  by  sliding  or  shear 31 

Influence  lines 31 

Actual  pressures  in  figures 34 

Haessler's  method 36 

Stepped  polygon 37 

Modified  equivalent  pressure  area  in  inclined  back  dam ...   37 

Curved  back  profiles 39 

Treatment  for  broken  line  profiles 41 

Example  of  Haessler's  method 42 

Relations  of  R.  N.  and  W 43 

Unusually  high  dams 43 

Pentagonal  profile  to  be  widened 47 

Silt  against  base  of  dam 50 

Ice  pressure ". 51 

Partial  overfall  dams.  .  .o  o  *.  j  j».» 52 


CONTENTS 

PAGE 

Notable  existing  dams 53 

Cheeseman  Lake  dam 53 

Analytical  check r  55 

Roosevelt  dam 56 

New  Croton  dam 58 

Assuan  dam 59 

Cross  River  and  Ashokan  dams 65 

Burrin  Juick  dam 65 

Arrow  Rock  dam 67 

Special  foundations 69 

Aprons  affect  uplift 70 

Rear  aprons  decrease  uplift 71 

Rock  below  gravel 72 

Gravity  dam  reinforced  against  ice  pressure 73 

Gravity  overfall  dams  or  weirs 75 

Characteristics  of  overfalls 75 

Approximate  crest  width 77 

Pressures  affected  by  varying  water  level 79 

Method  of  calculating  depth  of  overfall 82 

Objections  of  "Ogee"  overfalls 85 

Folsam  weir 85 

Dhukwa  weir 90 

Mariquina  weir 92 

Granite  Reef  weir 92 

Nira  weir 95 

Castlewood  weir 96 

American  dams  on  pervious  foundations 97 

Arched  dams 101 

Theoretical  and  practical  profiles 102 

Support'  of  vertical  water  loads  in  arched  dams 104 

Pathfinder  dam 104 

Shoshone  dam 107 

Sweetwater  dam 109 

Barossa  dam Ill 

Burrin  Juick  subsidiary  dam 112 

Dams  with  variable  radii 112 

Multiple  arch  or  hollow  arch  buttress  dams 113 

Multiple  arch  generally  more  useful  than  single  arch  dams. .  113 
Mir  Alam  dam. .  114 


CONTENTS 

PAGE 

Multiple  arch  or  hollow  arch  buttress  dams — continued 

Stresses  in  buttress 117 

Belubula  dam .118 

Ogden  dam 120 

Design  for  multiple  arch  dam 122 

Reverse  water  pressure , . .  124 

Pressure  on  foundations 125 

Flood  pressures 129 

Big  Bear  Valley  dam 131 

Hollow  slab  buttress  dams 136 

Formulas  for  reinforced  concrete 137 

Guayabal  dam 141 

Bassano  dam 146 

Submerged  weirs  founded  on  sand 151 

Percolation  beneath  dam 152 

Governing  factor  for  stability 153 

Vertical  obstruction  to  percolation 159 

Rear  apron 159 

Example  of  design  type  A 164 

Discussion  of  Narora  weir 167 

Sloping  apron  weirs,  type  B 169 

Restoration  of  Khanki  weir 171 

Merala  weir 171 

Porous  fore  aprons 173 

Okhla  and  Madaya  weirs 177 

Dehri  weir 178 

Laguna  weir 179 

Damietta  and  Rosetta  weirs 179 

Open  dams  or  barrages 182 

Barrage  defined 182 

Weir  sluices  of  Corbett  dam 189 

General  features  of  river  regulators 193 

Stability  of  Assiut  barrage 194 

Hindia  barrage 194 

American  vs.  Indian  treatment 196 

North  Mon  canal 200 

Upper  Coleroon  regulator.  . 201 

St.  Andrew's  Rapids  dam 201 

Automatic  dam  or  regulator 205 


INTRODUCTION 


A  N  unused  waterfall,  no  matter  how  beautiful,  appeals  to  an 
*••  engineer  mainly  as  an  economic  waste,  and  he  fairly  aches  to 
throw  a  dam  across  the  rushing  torrents  or  to  utilize  the  power  of 
the  water  which  glides  gracefully  over  the  falls  and  dashes  itself 
into  useless  spray  many  feet  below.  His  progress  in  the  past 
years  has,  however,  in  no  way  measured  up  to  his  desires,  but 
with  the  United  States  and  other  governments  undertaking  gigan- 
tic irrigation  projects  in  order  to  reclaim  vast  areas  of  tillable 
lands  and  with  hydroelectric  companies  acquiring  the  power 
rights  of  our  great  waterfalls,  the  last  few  years  have  witnessed 
wonderful  progress  in  this  type  of  engineering  work.  The  use  of 
reinforced  concrete  as  a  standard  material  and  the  solving  of  the 
many  problems  in  connection  with  it  has  greatly  simplified  and 
cheapened  the  construction,  thus  avoiding  the  greater  difficulties 
of  masonry  construction  usually  found  in  the  older  dams. 
9  All  of  this  progress  in  the  design  of  dams  and  weirs,  however, 
has  served  to  multiply  the  types  of  design  and  has  increased  the 
need  for  an  authoritative  and  up-to-date  treatise  on  the  theoretical 
and  practical  questions  involved.  The  author  of  this  work  has 
been  a  designing  engineer  for  more  than  a  generation  and  has 
built  dams  and  weirs  in  India,  Egypt,  Canada,  and  this  country. 
He  is,  therefore,  abundantly  qualified  to  speak,  not  only  from  the 
historic  side  of  the  work  but  from  the  modern  practical  side  as  well. 
In  addition  to  a  careful  analysis  of  each  different  type  of  profile, 
he  has  given  critical  studies  of  the  examples  of  this  type,  showing 
the  good  and  bad  points  of  the  designs.  A  wealth  of  practical 
problems  together  with  their  solution  makes  the  treatise  exceed- 
ingly valuable. 

9  It  is  the  hope  of  the  publishers  that  this  modern  treatise  will 
satisfy  the  demand  for  a  brief  but  authoritative  work  on  the  sub- 
ject and  that  it  will  find  a  real  place  in  Civil  Engineering  literature. 


DAMS  AND  WEIRS 

PART  I 


INTRODUCTION 

1 .  Definitions.    A  dam  may  be  defined  as  an  ijnperidoiis  wall 
of  masonry,  concrete,  earth,  or  loose  rock  which  upholds  a  mass  of 
water  at  its  rear,  while  its  face  or  lower  side  is  free  from  the  pressure 
of  water  to  any  appreciable  extent.     The  waste  water  of  the  reser- 
voir impounded  by  the  dam  is  disposed  of  by  means  of  a  waste  weir, 
or  by  a  spillway  clear  of  the  wrork,  or  in  rare  cases,  by  sluice  openings 
in  the  body  of  the  dam. 

Weirs,  or  overfall  dams,  although  often  confounded  with  bulk- 
head dams,  differ  from  the  latter  in  the  following  points,  first, 
that  the  water  overflows  the  crest,  and  second,  that  the  tail  water 
is  formed  below  the  dam.  These  two  differences  often  modify  the 
conditions  of  stress  which  are  applicable  in  the  design  of  dams 
proper,  and  consequently  the  subject  of  weirs  demands  separate 
treatment. 

2.  Classification.     Dams    and    weirs    may    be    classified   as 
follows : 

1.  Gravity  Dams 

2.  Gravity  Overfalls,  or  Weirs 

3.  Arched  Dams 

4.  Hollow  Arch  Buttress  Dams 

5.  Hollow  Slab  Buttress  Dams 

6.  Submerged  Weirs 

7.  Open  Dams,  or  Barrages 

The  subjects  of  earth,  rock  fill,  and  steel  dams  will  not  be 
treated  in  this  article,  as  the  matter  has  been  already  dealt  with  in 
other  volumes.  Graphical  as  well  as  analytical  methods  will  be 
made  use  of,  the  former  procedure  being  explained  in  detail  as 
occasion  demands. 


DAMS  AND  WEIRS 


GRAVITY  DAMS 

GENERAL  DISCUSSION  OF  DAMS 

A  gravity  dam  is  one  in  which  the  pressure  of  the  water  is 
resisted  by  the  weight  or  "gravity"  of  the  dam  itself. 

3.  Pressure  of  Water  on  Wall.  The  hydrostatic  pressure  of 
the  water  impounded  by  a  wall  or  dam  may  be  graphically  repre- 
sented by  the  area  of  a  triangle  with  its  apex  at  the  surface  and  its 
base  drawn  normal  to  the  back  line  of  the  dam,  which  base  is  equal 
or  proportionate  to  the  vertical  depth.  When  the  back  of  the  wall 

H2 
is  vertical,  as  in  Fig.  1,  the  area  of  this  pressure  triangle  will  be  — 

H  being  the  vertical  height.    When,  as  in  Fig.  2,  the  back  is  inclined, 

TTf  TT 

this  area  will  be  — - — ,  H'  being  the  inclined  length  of  the  exposed 

surface,  which  equals  H  cosec  $. 

The  actual  pressure  of  water  per  unit  length  of  dam  is  the 
above  area  multiplied  by  the  unit  weight  of  water.  This  unit 


Fig.  1.     Water  Pressure  Area 
with  Back  of  Dam  Vertical 


Fig.  2.     Water  Pressure  Area  with 
Back  of  Dam  Inclined 


weight  is  symbolized  by  w,  which  is  62.5  pounds,  or  $5  short  ton,  or 
sV  long  ton,  per  cubic  foot. 

Unit  Pressure.  The  pressure  per  square  foot,  or  unit  pressure 
on  the  wall  at  any  point,  is  measured  by  the  corresponding  ordinate 
of  the  above  triangle,  drawn  parallel  to  its  base,  and  is  evidently 
the  same  in  both  Figs.  1  and  2.  The  total  pressure  on  the  inclined 
back  as  represented  by  the  triangle  in  Fig.  2  will,  however,  be 
greater  than  in  Fig.  1. 

4.  Method  for  Graphical  Calculations.  For  graphical  calcula- 
tions when  forces  of  dissimilar  unit  weight  or  specific  gravity  are 


DAMS  AND  WEIRS  3 

engaged,  as  in  the  case  of  water  and  masonry,  or  earth  and  masonry, 
it  is  the  usual  practice  to  reduce  them  to  one  common  denominator 
by  making  alterations  in  the  areas  of  one  or  the  other,  the  weight  of 
the  masonry  being  usually  taken  as  a  standard.  This  result  is 
effected  by  making  the  bases  of  the  triangles  of  water  pressure 

TT 

equal,  not  to  H,  but  to  — ,  p  (rho)  being  the  sign  of  the  specific 

P 

gravity  of  the  solid  material  in  the  wall.  The  triangle  thus  reduced 
will  then  represent  a  weight  or  area  of  masonry  1  unit  thick,  equiva- 
lent to  that  of  water.  This  device  enables  the  item  of  unit  weight, 
which  is  wXp  to  be  eliminated  as  a  common  factor  from  the  forces 
engaged,  i.  e.,  of  the  water  pressure  and  of  the  weight  of  the  wall. 
The  factor  thus  omitted  has  to  be  multiplied  in  again  at  the  close 
of  the  graphical  operation,  only,  however,  in  cases  where  actual 
pressures  in  tons  or  pounds  are  required  to  be  known. 

Value  of  p.  The  values  ordinarily  adopted  for  p,  the  specific 
gravity  of  masonry  or  concrete,  are  2\  and  2.4,  i.e. , equivalent  to 
weights  of  141  and  150  pounds,  respectively,  per  cubic  foot,  while 
for  brickwork  2  is  a  sufficiently  large  value.  The  value  of  wp  in  the 

o 

former  case  will  be  .069  ton  and  in  the  latter  .075,  or  —  ton. 

In  some  cases  the  actual  value  of  p  mounts  as  high  as  2.5  and 
even  2.7,  when  heavy  granite  or  basalt  is  the  material  employed. 

The  reduction  thus  made  in  the  water  pressure  areas  has  further 
the  convenience  of  reducing  the  space  occupied  by  the  diagram. 
The  areas  of  the  reduced  triangles  of  water  pressure  in  Figs.  1  and 

H*      A  H'H  .    . 

2  are  — —  and  — - —  respectively. 
2p  2p 

5.  Conditions  of  "Middle  Third"  and  Limiting  Stress.  Sec- 
tions of  gravity  dams  are  designed  on  the  well-known  principle  of 
the  "middle  third."  This  expression  signifies  that  the  profile  of 
the  wall  must  be  such  that  the  resultant  pressure  lines  or  centers  of 
pressure  due  first  to  the  weight  of  the  dam  considered  alone,  and 
second  with  the  external  water  pressure  in  addition,  must  both 
fall  within  the  middle  third  of  the  section  on  any  horizontal  base. 
These  two  conditions  of  stress  are  designated,  Reservoir  Empty 
(R.E.)  and  Reservoir  Full  (R.F.).  The  fulfillment  of  this  condition 
insures  the  following  requirement:  The  maximum  compressive  ver- 


4  DAMS  AND  WEIRS 

tical  unit  stress  (s) ,  or  reaction  on  the  base  of  a  flam,  shall  not  exceed 
twice  the  mean  compressive  unit  stress,  or,  stated  symbolically, 

8^2*1 

Now  the  mean  vertical  compressive  unit  stress  Si  is  the  weight  of  the 
structure  divided  by  its  base  length-  i.e., 

W 
Si  =  J 

2W 

Hence,  s,  the  maximum  vertical  unit  pressure,  should  not  exceed-; — 

b 

Further  comments  on  the  distribution  of  the  reaction  on  the  base 
of  a  dam  will  be  made  in  a  later  paragraph. 

6.  Compressive  Stress  Limit.  A  second  condition  imposed  is 
that  of  the  internal  compressive  stress  limit,  that  is :  The  maximum 
permissible  compressive  unit  stress  which  is  developed  in  the  interior 
of  the  masonry  of  the  dam,  must  not  be  exceeded.  This  value  can  be 
experimentally  found  by  crushing  a  cube  of  the  material  employed, 
and  using  a  factor  of  safety  of  6  or  8.  Cement  concrete  will  crush 
at  about  2000  pounds  per  square  inch,  equivalent  to  144  tons  (of 
2000  pounds)  per  square  foot.  The  safe  value  of  s  would  then  be 

1   A  A 

tons   per   square   foot.     For   ordinary   lime   concrete    as 


employed  in  the  East,  the  limit  pressure  adopted  is  generally  8 
"long"  tons,  equivalent  to  9  tons  of  2000  pounds.  Ten  "long" 
tons,  or  11.2  "short"  tons  is  also  a  common  value. 

DESIGN  OF  DAMS 

7.  Theoretical  Profile.  The  theoretically  correct  profile  of  a 
so-termed  "low"  masonry  dam,  i.e.,  one  of  such  height  that  the  limit 
stress  is  not  attained  under  the  conditions  above  outlined,  is  that 
of  a  right-angled  triangle  having  its  back  toward  the  water  vertical, 
and  its  apex  at  the  water  surface.  It  can  be  proved  that  the  proper 
base  width  b  of  this  triangle  is  expressed  by  the  formula 

(i) 

This  profile,  shown  in  Fig.  3,  will  be  termed  the  "elementary  trian- 
gular profile",  as  on  it  the  design  of  all  profiles  of  dams  is  more  or 
less  based.  In  this  expression,  H  is  the  vertical  height.  The  base 


DAMS  AND  WEIRS  5 

TT 

width  of  —j=r  insures  the  exact  incidence  of  the  vertical  resultant  (IF) 
Vp 

(R.E.)  and  of  the  inclined  resultant  R  (R.F.)  at  the  inner  and  outer 
edge,  respectively,  of  the  central  third  division  of  the  base.  The 
condition  of  the  middle  third  is  thus  fulfilled  in  the  most  economical 
manner  possible,  a  factor  of  safety  of  2  against  overturning  is 
obtained,  and  further,  the  angle  of  inclination  of  the  resultant  R 
with  regard  to  the  base  is  usually  such  as  to  preclude  danger  of 
failure  by  sliding. 

The  fore  slope  or  hypothenuse  will  be  in  the  ratio  1 :  V p  which, 
when  p  =  2J,  will  equal   2:3,   a   slope   very  commonly  adopted, 


DOUBLE  SCRLE 


Fig.  3.     Elementary  Triangular  Profile  for  "Low"  Masonry  Dam 

and  with  p  =  2.4  the  ratio  will  be  1:1.549.    The  area  of  the  ele- 

H* 

mentary  triangle  is  — =•  while,  as  we  have  seen,  that  of  the  water 

2VP 

H2 

pressure  is  •— .    6  is  the  vertical  angle  between  W  and  R,  and 

_p 

sec  6  --£±1—1.187  with  p  =  2.4. 
Vp 

In  Fig.  3  the  resultant  pressure  lines  are  drawn  to  intersect  the 
base  so  as  to  afford  ocular  proof  of  the  stability  of  the  section  under 
the  postulated  conditions. 

Graphical  Method.  The  graphical  procedure  will  now  be  briefly 
explained,  and  also  in  the  future  as  fresh  developments  arise,  for 


6  DAMS  AND  WEIRS 

the  benefit  of  those  who  are  imperfectly  acquainted  with  this  valu- 
able labor-saving  method. 

There  are  two  forces  engaged,  P  the  horizontal,  or,  it  may  be  PI, 
the  inclined  water  pressure  acting  through  the  center  of  gravity  of  its 
area  normal  to  the  back  of  the  wall,  and  W  the  weight  or  area  of  the 
wall.  Of  these  two  forces  the  item  wp,  or  unit  weight,  has  already 
been  eliminated  as  a  common  factor,  leaving  the  pressures  repre- 
sented by  superficial  areas.  As,  however,  the  height  H  is  also  com- 
mon to  both  triangles,  this  can  likewise  be  eliminated.  The  forces 
may  then  be  represented  simply  by  the  half  widths  of  the  triangular 
areas  by  which  means  all  figuring  and  scaling  may  be  avoided. 

First,  a  force  polygon  has  to  be  constructed.  In  Fig.  3a,  P  is 
first  drawn  horizontally  to  designate  the  water  pressure,  its  length 
being  made  equal  to  the  half  width  of  its  pressure  area  in  Fig.  3. 
From  the  extremity  of  P,  the  load  line  W  is  drawn  vertically,  equal 
to  the  half  width  of  the  elementary  triangular  profile,  then  the 
closing  line  R  according  to  the  law  of  the  triangle  of  forces  will 
represent  the  resultant  in  magnitude  and  direction.  Second,  the 
lines  of  actual  pressures  reciprocal  to  those  on  the  force  polygon 
will  have  to  be  transferred  to  the  profile.  The  incidence  of  the 
resultant  water  pressure  on  the  back  is  that  of  a  line  drawn  through 

TT 

the  e.g.  of  the  area  of  pressure,  parallel  to  its  base,  in  this  case,  at  — , 

o 

or  one-third  the  height  of  the  water-pressure  triangle,  above  the 
base.  Its  direction,  like  that  of  the  base,  is  normal  to  the  back,  in 
this  case  horizontal,  and  if  prolonged  it  will  intersect  the  vertical 
force  W,  which  in  like  manner  acts  through  the  center  of  gravity  of 
the  elementary  profile  of  the  wall.  From  this  point  of  intersection 
the  resultant  R  is  drawn  parallel  to  its  reciprocal  in  Fig.  3a.  Both 
W  and  R  are  continued  until  they  cut  the  base  line,  and  these  points 
of  intersection  will  be  found  to  be  exactly  at  the  inner  and  outer 
edges  of  the  middle  third  division  of  the  base.  It  will  be  seen  that 
when  the  reservoir  is  empty  the  center  of  pressure  on  the  base  is  at 
the  incidence  of  W,  when  full  it  is  shifted  to  that  of  R. 

Analytical  Method.  The  same  proof  can  be  made  analytically 
as  follows :  The  weight  of  the  two  triangles  W  and  P  can  be  repre- 

TT  TT 

sented  by  their  bases  which  are  —:=  and  — ,  respectively.  If  moments 

Vp  P 


DAMS  AND  WEIRS  •      7 

be  taken  about  the  outer  edge  of  the  middle  third,  the  lever  arm  of 
the  vertical  force  W  is  clearly  —  or  — j=  and  that  of  P,  the  horizontal 

O  O  \  n 

force,  is  the  distance  of  the  center  of  gravity  of  the  triangle  of  water 

TJ 

pressure  above  the  base,  viz,  — .     The  equation  will  then  stand 


or 


_ 

3p      3p 

If  the  actual  values  of  R  and  of  W  were  required,  their  measured  or 
calculated  lengths  would  have  to  be  multiplied  by  H  and  by  wp  in  order 
to  convert  them  to  tons,  pounds,  or  kilograms,  as  may  be  required. 
In  many,  in  fact  most,  cases  actual  pressures  are  not  required  to  be 
known,  only  the  position  of  the  centers  of  pressures  in  the  profile. 
Thus  a  line  of  pressures  can  be  traced  through  a  profile  giving 
the  positions  of  the  centers  of  pressure  without  the  necessity  of 
converting  the  measured  lengths  into  actual  quantities.  In  the 
elementary  triangle,  Fig.  3,  the  value  of  the  vertical  resultant  W  is 

.          p .    That  of  R  required  in  the  older  methods  of  calculation 


is 


The  following  values  relative  to  p  will  be  found  useful. 


P    OR 

SPECIFIC 
GRAVITY 

VP 

l 

VP 

j_ 
P 

POUNDS  PER 
CUBIC  FOOT 

TONS  PER 
CUBIC  FOOT 

2 

1.414 

.71 

.5 

125 

.0625 

2i 

1.5 

I 

1 

141 

.07 

2.4 

1.55 

.645 

.417 

150 

.075  =  A 

2.5 

1.58 

.633 

.4 

156 

.078 

2.7 

1.643 

.609 

.37 

168.7 

.084 

Profile  with  Back  Inclined.  If  the  elementary  profile  be  canted 
forward  so  that  its  back  is  inclined  to  the  vertical,  it  will  be  found 
that  the  incidence  of  R  will  fall  outside  the  middle  third  while  that 
of  W  will  be  inside.  The  base  will,  therefore,  have  to  be  increased 

H 

above  -=-. 


s 


DAMS  AND  WEIRS 


When  the  back  is  overhanging,  on  the  other  hand,  R  will  fall 
inside  and  W  outside  the  middle  third.  The  vertically  backed 
section  is  consequently  the  most  economical. 

8.  Practical  Profile.  In  actual  practice  a  dam  profile  must  be 
provided  with  a  crest  of  definite  width,  and  not  terminate  in  the 
apex  of  a  triangle.  The  upper  part  of  a  dam  is  subjected  to  indefi- 
nite but  considerable  stresses  of  an  abnormal  character,  due  to  extreme 
changes  in  temperature,  consequently  a  solid  crest  is  a  necessity. 
The  imposition  of  a  rectangular  crest,  as  shown  in  dotted  lines  on 


Fig.  4. 


Practical  Pentagonal  Profile  for  "Low' 
Masonry  Dam 


Fig.  3,  transforms  the  triangular  profile  into  a  pentagon.  This  has 
the  effect  of  increasing  the  stability  of  the  section  (R.F.)  so  that 
the  base  width  can  be  somewhat  reduced,  at  the  same  time  the 
vertical  resultant  W  (R.E.),  falls  outside  the  middle  third,  but  to 
so  small  an  extent  that  this  infringement  of  the  imposed  condition 
is  often  entirely  neglected.  In  order  to  provide  against  this,  a  strip 
of  material  will  have  to  be  added  to  the  back  of  the  plain  pentagonal 
profile.  Fig.  4  is  a  diagram  explanatory  of  these  modifications. 
The  dimensions  of  this  added  strip,  as  well  as  its  position,  can  be 


DAMS  AND  WEIRS 


9 


conveniently  expressed  in  terms  of  (k)  the  crest  width — i.e.,  AB 
in  Fig.  4.  The  line  of  pressure  (R.E.)  will  begin  to  leave  the  middle 
third  at  the  depth  AD,  which  is  found  by  calculation  which  need 
not  be  produced  here,  to  be  2&Vp.  Below  the  point  D,  the 
divergence  of  the  line  of  pressure  will  continue  for  a  further  depth 
DE,  the  point  E,  being  close  upon  3.1&Vp  below  the  crest,  or 
l.l&Vp  below  D.  Below  point  E,  the  line  of  pressure  will  no 
longer  diverge  outward,  but  will  tend  to  regain  its  original  position, 
consequently  no  further  widening  will  be  necessary,  and  the  added 


El.  16/0,3 


ELI446.S 


Fig.  5.     Profile  of  Chartrain  Dam  Showing  Crest  with  Overhang 

strip  will  be  rectangular  in  form  down  to  the  base.  The  points  D 
and  F  being  joined,  this  portion  of  the  back  will  be  battered.  The 
width  of  this  added  strip  EF  will  be,  with  close  approximation, 

^  or  ML 
lo 

9.  Crest  Width.  The  crest  width  of  a  dam  should  be  propor- 
tioned to  its  actual  height  in*  case  of  a  "low"  dam,  and  in  the  case  of 
a  "high"  dam  to  the  limiting  height — i.e.,  to  that  depth  measured 
below  the  crest  at  which  the  maximum  stress  in  the  masonry  is  first 


10 


DAMS  AND  WEIRS 


reached.  Thus  in  "high"  dams  the  upper  part  can  always  be  of 
the  same  dimensions  except.  w,here  the  requirements  of  cross  com- 
munication necessitate  a  wider  crest. 

The  effect  of  an  abnormally  wide  crest  can  be  modified  by 
causing  it  to  overhang  the  fore  slope,  this  widening  being  •  carried 
by  piers  and  arches.  A  good  example  of  this  construction  occurs 
in  the  Chartrain  dam,  Fig.  5.  The  arches  form  a  stiff  but  light 
finish  to  the  dam  and  have  a  pleasing  architectural  effect.  The 
same  procedure,  'but  in  a  less  pronounced  degree,  is  carried  out  in 

the  Croton  dam,  Fig.  27. 

The  formula  for  crest 
width  can  be  expressed 
either  in  terms  of  the  limit- 
ing height  HI,  or  of  the 
base  b,  where  the  limiting 
height  is  not  attained,  and 
a  good  proportion  is  given 
by  the  following  empirical 
rule: 

(2) 


Fig.  6.     Pentagonal  Profile — Back  Vertical 


or 

A;  =  .156        (2a) 

This  latter  formula  makes 

the  crest  width  a  function  of  the  specific  gravity  as  well  as  of  the 
height,  which  is  theoretically  sound. 

10.  Rear  Widening.  Where  the  rear  widening  of  a  "low" 
dam  is  neglected  or  where  a  uniform  batter  is  substituted  for  the 
arrangement  shown  in  Fig.  4,  the  profile  will  be  pentagonal  in  out- 
line. When  the  back  is  vertical  the  two  triangles  composing  the 
body  of  the  dam  are  similar.  If  the  ratio  existing  between  the  crest 

k 

(k)  and  the  base  (6),  or  —  be  designated  by  r,  then  k  =  br,  and  h,  the 
o 

depth  of  the  vertical  side  in  Fig.  6,  =  Hr  and  kXh=  Hbr*. 

In  order  to.  find  what  value  the  base  width  b  should  have,  so 
that  the  center  of  pressure  (R.F.)  will  fall  exactly  at  the  edge  of  the 
middle  third,  the  moments  of  all  the  forces  engaged  will  have  to  be 
taken  about  this  point  and  equated  to  zero.  The  vertical  forces 


DAMS  AND  WEIRS  11 

consist  of  W,  the  lower,  and  Wi  the  upper  triangle;  the  horizontal 
of  P,  the  water  pressure. 

1  1.  Method  of  Calculation.  The  pressures  can  be  represented 
by  the  areas  of  the  prisms  involved,  the  triangle  of  water  pressure 
being  as  usual  reduced  by  dividing  its  base  by  p.  A  further  elim- 

ination of  common  factors  can  be  achieved  by  discarding  —  which 
is  common  to  all  three  forces,  the  area  W\  being  represented  by  br2 

ZjTZ      9 

because  the  actual  original  value  is  -  .     The  forces  then  are  W, 

£t 

represented  by  b;  W\  by  6r2;  and  P  by  —  ;  the  actual  value  of  the 

'  P 

//2 

latter  being  —  —  .     The  lever  arm  distances  of  the  c.g.'s  of  these  three 
Zp 

forces  from  A,  the  incidence  of  R,  are  as  follows:  of  W,  —  ,  of  PFi, 

o 

2  H  1 

—  (b  —  br),  and  of  P,  —  .   The  equation  will  then  stand,  eliminating  -, 

o  o  o 

-—  =  Q 

P 
or 

62(l+2r2-2r3)-—  =  0 

P 
whence 


.. 

p     Vl+2r2-2r3 


(3) 


The  value  of  b  thus  obtained  will  prove  a  useful  guide  in  deciding 
the  base  width  even  when  the  back  of  the  wall  is  not  vertical,  as  only 
a  small  increase  will  be  needed  to  allow  for  the  altered  profile.  When 

k  1 

—or  r  =  .15  the  reducing  coefficient  works  out  to         _,  the  reciprocal 

of  which  is  .981.     Thus  with  a  profile  80  feet  high  with  p  =  2.5  and 

on 

r  =  .15,  the  base  width  of  the  pentagonal  profile  will  be  b  =    _  X  .981 

V2.5 

=  49.64  feet;  the  decrease  in  base  width  below  that  of  the  elementary 
profile  without  crest  will  be  50.60-49.64  =  0.96  feet.  The  crest 
width  will  be  49.64  X.  15  =  7.45  feet.  In  actual  practice,  the  dimen- 
sions would  be  in  round  numbers,  50  feet  base  and  1\  feet  crest 


12 


DAMS  AND  WEIRS 


width  as  made  on  Fig.  6.  The  face  of  the  profile  in  Fig.  6  is  made 
by  joining  the  toe  of  the  base  with  the  apex  of  the  triangle  of  water 
pressure. 

Graphical  Process.  The  graphical  processes  of  finding  the 
incidences  of  W  and  of  R  on  the  base  are  self-explanatory  and  are 
shown  on  Fig.  6.  The  profile  is  divided  into  two  triangular  areas, 
(1),  45  square  feet  and  (2),  2000  square  feet.  The  two  final  result- 
ants fall  almost  exactly  at  the  middle  third  boundaries,  W,  as 
might  be  conjectured,  a  trifle  outside.  Areas  are  taken  instead  of 
|  widths,  owing  to  //  not  being  a  common  factor. 

Analytical  Process.  The  analytical  process  of  taking  moments 
about  the  heel  is  shown  below: 


AREA 

LEVER  ARM 

MOMENT 

m 

4^ 

7.5X2 

OOK 

w 
(2) 

2000 

3 
50 
3 

AAi) 
33333 

W 

2045 

33558 

The  value  33,558,  which  is  the  total  moment  of  parts  equals  the 
moment  of  the  whole  about  the  same  point  or 

2045X^  =  33558 

3  =  16.41  feet 

50 

The  incidence  of  W  is  therefore-  --  16.41  =  .26  ft.  outside  the  middle 

o 

third.     To  find  that  of  R  relative  to  the  heel,  the  distance  (see 

PH     1280X80     102400 
section  17)  between  W  and  R  is          =  = 


The  distance  of  R  from  the  heel  is  therefore  16.69+16.41  =  33.10  ft. 

2 
The  —  point  is  33.33  feet  distant,  consequently  the  incidence  of  R 

o 

is  .23  foot  within  the  middle  third. 

If  the  base  and  crest  had  been  made  of  the  exact  dimensions 
deduced  from  the  formula,  the  incidence  of  R  would  be  exactly  at 

2 

the  —  point  while  W  would  fall  slightly  outside  the  one-third  point. 
o 


DAMS  AND  WEIRS 


13 


12.  Variation  of  Height.    The  height  of  a  dam  is  seldom  uni- 
form throughout;  it  must  vary  with  the  irregularities  of  the  river 
bed,  so  that  the  maximum  section  extends  for  a  short  length  only, 
while  the  remainder  is  of  varying  height..    This  situation  will  affect 
the  relationship  between  the  crest  width  and  the  height,  and  also 
the  base  width.    To  be  consistent,  the  former  should  vary  in  width 
in  proportion  to  the  height.    This,  however,  is  hardly  practicable, 
consequently  the  width  of  the  crest  should  be  based  more  on  the 
average  than  on  the  maximum  height,  and  could  be  made  wider 
wherever  a  dip  occurs  in  the  foundation  level. 

13.  High  and  Wide  Crest.     In  case  of  a  very  high  as  well  as 
wide  crest,  i.e.,  one  carried  much 

higher  than  the  apex  of  the  trian- 
gle of  water  pressure,  it  is  not 
desirable  to  reduce  the  base  width 
much  below  that  of  the  elemen- 
tary triangle.  The  excess  of 
material  in  the  upper  quarter  of 
a  "low"  dam  can  be  reduced  by 
manipulating  the  fore  slope. 
This  latter,  which  is  drawn  up- 
ward from  the  toe  of  the  base, 
in  Fig.  7,  can  be  aligned  in  three 
directions.  First,  by  a  line  ter- 
minating at  the  apex  of  the  trian- 
gle of  water  pressure;  second,  it 
can  be  made  parallel  to  that  of 
the  elementary  profile,  that  is,  it  can  be  given  an  inclination  of 

—=  to  the  vertical,  and  third,  the  slope  or  batter  can   be  made 

VP 

flatter  than  the  last.  This  latter  disposition  is  only  suitable  with 
an  abnormally  high  and  wide  crest  and  is  practically  carried  out 
in  the  Chartrain  dam,  Fig.  5,  where  the  base  is  not  reduced  at  all 

TT 

below  -7=. 

VP 

Reduction  to  any  large  extent,  of  the  neck  of  the  profile  thus 
effected  is,  however,  not  to  be  commended,  as  the  upper  quarter  of 
a  dam  is  exposed  to  severe  though  indeterminate  stresses  due  to 


B  B, 
Fig.  7. 


C  B  OR  C,B,  =50.0' 


Profile  Showing  Different  Disposition 
of  Fore  Slope 


14  DAMS  AND  WEIRS 

changes  of  temperature,  wind  pressure,  etc.,  and  also  probably  to 
masses  of  ice  put  in  motion  by  the  wind.  The  Cross  River  dam, 
to  be  illustrated  later,  as  well  as  the  Ashokan  dam,  are  examples  of 
an  abnormally  thick  upper  quarter  being  provided  on  account  of 
ice.  Whatever  disposition  of  the  fore  slope  is  adopted,  the  profile 
should  be  tested  graphically  or  analytically,  the  line  of  pressure,  if 
necessary,  being  drawn  through  the  profile,  as  will  later  be  explained. 

From  the  above  remarks  it  will  be  gathered  that  the  design  of 
the  section  of  a  dam  down  to  the  limiting  depth  can  be  drawn  by  a 
few  lines  based  on  the  elementary  profile  which,  if  necessary,  can 
be  modified  by  applying  the  test  of  ascertaining  the  exact  position 
of  the  centers  of  pressure  on  the  base.  If  the  incidence  of  these 
resultants  falls  at  or  close  within  the  edge  of  the  middle  third  divi- 
sion of  the  base,  the  section  can  be  pronounced  satisfactory;  if 
otherwise,  it  can  easily  be  altered  to  produce  the  desired  result. 

Freeboard.  The  crest  has  to  be  raised  above  actual  full  reservoir 
level  by  an  extent  equal  to  the  calculated  depth  of  water  passing 
over  the  waste  weir  or  through  the  spillway,  as  the  case  may  be. 
This  extra  freeboard,  which  adds  considerably  to  the  cost  of  a 
work,  particularly  when  the  dam  is  of  great  length  and  connected 
with  long  embankments,  can  be  avoided  by  the  adoption  of  auto- 
matic waste  gates  by  which  means  full  reservoir  level  and  high  flood 
level  are  merged  into  one. 

In  addition  to  the  above,  allowance  is  made  for  wave  action, 
the  height  of  which  is  obtained  by  the  following  formula: 

hw  =  l.^F+(2.5-^Y)  (4) 

In  this  expression  F  is  the  "fetch",  or  longest  line  of  exposure  of 
the  water  surface  to  wTind  action  in  miles.  Thus  if  F  =  4  miles,  the 
extra  height  required  over  and  above  maximum  flood  level  will  be 
(1.5X2)  +  (2.5-1.4)=4.1  feet.  If  F=W  miles,  hw  will  work  out  to 
5|  feet.  The  apex  of  the  triangle  of  water  pressure  must  be  placed 
at  this  higher  level;  the  crest,  however,  is  frequently  raised  still 
higher,  so  as  to  prevent  the  possibility  of  water  washing  over  it. 

14.  Example.  The  working  out  of  an  actual  example  under 
assumed  conditions  will  now  be  given  by  both  graphical  and  analyti- 
cal methods.  Fig.  8  represents  a  profile  50  feet  in  height  with  crest 
level  corresponding  with  the  apex  of  the  triangle  of  water  pressure. 


DAMS  AND  WEIRS 


15 


The  assumed  value  of  p  is  2J.     The  outline  is  nearly  pentagonal, 

H     2 

the  crest  width  is  made  .156  and  the  base  width  is  the  full  - 

Vp     3 

X  50  =  33.3  feet,  the  crest  width  is  thus  5  feet.  The  back  slope  is 
carried  down  vertically  to  the  point  e,  a  distance  of  8  feet,  and  from 
here  on,  it  is  given  a  batter  of  1  in  50.  The  outset  at  the  heel  beyond 
the  axis  of  the  dam,  which  is  a  vertical  line  drawn  through  the  rear 


(b) 

Fig.  8.     Diagram  Showing  Suitable  Profile  for  50-Foot  Dam 

edge  of  the  crest  is  therefore  .84  foot.  The  toe  is  set  in  the  same 
extent  that  the  heel  is  set  out.  The  face  line  of  the  body  is  formed 
by  a  line  joining  the  toe  with  the  apex  of  the  water-pressure  triangle. 
If  the  face  line  were  drawn  parallel  to  the  hypothenuse  of  the  ele- 
mentary triangle,  i.e.,  to  a  slope  of  1  :  Vp,  it  would  cut  off  too  much 
material,  the  area  of  the  wall  being  then  but  very  little  in  excess  of 
that  of  the  elementary  triangle,  which,  of  course,  is  a  minimum 
quantity.  As  will  be  seen  later,  the  analysis  of  the  section  will  show 
that  the  adopted  base  width  could  have  been  reduced  below  what 


16  DAMS  AND  WEIRS 

has  been  provided,  to  an  extent  somewhat  in  excess  of  that  given  in 
formula  (3). 

15.  Graphical  Method.  The  graphical  procedure  of  drawing 
the  resultant  lines  W  (R.E.)  and  R  (R.F.)  to  their  intersection  of  the 
base  presents  a  few  differences,  from  that  described  in  section  7, 
page  6,  with  regard  to  Fig.  3.  Here  the  profile  is  necessarily  divided 
into  two  parts,  the  rectangular  crest  and  the  trapezoidal  body.  As 
the  three  areas  (1),  (2),  and  PI,  are  not  of  equal  height,  the  item  // 
cannot  be  eliminated  as  a  common  factor,  consequently  the  forces 
will  have  to  be  represented  as  in  Fig.  6  by  their  actual  superficial 
areas,  not  by  the  half  width  of  these  areas  as  was  previously  the  case. 
In  Fig.  8a  the  vertical  load  line  consists  of  the  areas  1  and  2  totaling 
844  square  feet,  which  form  W.  The  water  pressure  PI  is  the  area 

TT 

of  the  inclined  triangle  whose  base  is  — .    This  is  best  set  out  graph- 

P 

ically  in  the  force  polygon  by  the  horizontal  line  P,  made  equal  to 

,     ,     .  ,.  ,    .     H2     2500X2 

the  horizontal  water  pressure,  which  is  —  = =  555  square 

Zp 

feet.  The  water-pressure  area  strictly  consists  of  two  parts  corre- 
sponding in  depth  to  (1)  and  (2)  as  the  upper  part  is  vertical,  not 
inclined,  but  the  difference  is  so  slight  as  to  be  inappreciable,  and 
so  the  area  of  water  pressure  is  considered  as  it  would  be  if  the  back  of 
the  wall  were  in  one  inclined  plane.  In  Fig.  8  the  line  PI  normal  to 
the  back  of  the  wall  is  drawn  from  the  point  of  origin  0  and  it  is  cut 
off  by  a  vertical  through  the  extremity  of  the  horizontal  line  P. 
This  intercepted  length  0  0\  is  clearly  the  representative  value  of 
the  resultant  water  pressure,  and  the  line  joining  this  point  with  the 
base  of  the  load  line  W  is  R,  the  resultant  of  W  and  of  PI.  If  a 
horizontal  line  A  B  be  drawn  from  the  lower  end  of  the  load  line  W 
it  will  cut  off  an  intercept  ( N)  from  a  vertical  drawn  through  the 
termination  of  PI.  This  line  AB  =  P,  and  N  is  the  vertical  com- 
ponent of  R,  the  latter  being  the  resultant  of  W  and  PI  as  well  as 
of  N  and  P.  When  the  back  is  vertical,  N  and  W  are  naturally 
identical  in  value,  their  difference  being  the  weight  of  water  over- 
lying the  inclined  rear  slope. 

The  further  procedure  consists  in  drawing  the  reciprocals  of 
the  three  forces  PI,  IF,  and  R  on  the  profile.  The  first  step  consists 
in  finding  the  centers  of  gravity  of  the  vertical  forces  1  and  2  in  which 


DAMS  AND  WEIRS  17 

the  hexagonal  profile  is  divided.  That  of  (1)  lies  clearly  in  the 
middle  of  the  rectangle  whose  base  is  de.  The  lower  division  (2)  is 
a  trapezoid.  The  center  of  gravity  of  a  trapezoid  is  best  found  by 
the  following  extremely  simple  graphical  process.  From  d  draw 
dh  horizontally  equal  to  the  base  of  the  trapezoid  fg  and  from  g, 
gj  is  set  off  equal  to  de;  join  hj,  then  its  intersection  with  the  middle 
line  of  the  trapezoid  gives  the  exact  position  of  its  center  of  gravity. 
Thus  a  few  lines  effect  graphically  what  would  involve  considerable 
calculation  by  analytical  methods,  as  will  be  shown  later. 

The  next  step  is  to  find  the  combined  c.  g.  of  the  two  parallel 
and  vertical  forces  1  and  2.  To  effect  this  for  any  number  of  parallel 
or  non-parallel  forces,  two  diagrams  are  required,  first,  a  so-termed 
force  and  ray  polygon  and,  second,  its  reciprocal,  the  force  and 
chord,  or  funicular  polygon.  The  load  line  in  Fig.  8a  can  be  utilized 
in  the  former  of  these  figures.  First,  a  point  of  origin  or  nucleus  of 
rays  must  be  taken.  Its  position  can  be  anywhere  relative  to  the 
load  line,  a  central  position  on  either  side  being  the  best.  The 
point  Oi ,  which  is  the  real  origin  of  the  force  polygon  at  the  extremity 
of  PI  can  be  adopted  as  nucleus  and  often  is  so  utilized,  in  which 
case  the  force  line  PI  and  R  can  be  used  as  rays,  only  one  additional 
ray  being  required.  For  the  sake  of  illustration,  both  positions  for 
nucleus  have  been  adopted,  thus  forming  twTo  force  and  ray  poly- 
gons, both  based  on  the  same  load  line,  and  two  funicular  polygons, 
the  resultants  of  which  are  identical.  The  force  and  ray  polygon  is 
formed  by  connecting  all  the  points  on  the  load  line  with  the  nucleus 
as  shown  by  the  dotted  line  a,  b,  and  c,  and  a' ',  b' ',  and  c1 '.  Among  the 
former,  a  and  c  are  the  force  lines  PI  and  R,  the  third,  b,  joins  the 
termination  of  force  (1)  on  the  load  line  with  the  nucleus.  These 
lines  a,  b,  c,  are  the  rays  of  the  polygon.  Having  formed  the  force 
and  ray  diagram,  in  order  to  construct  the  reciprocal  funicular 
polygon  86  the  force  lines  (1)  and  (2)  on  the  profile  Fig.  8  are  con- 
tinued down  below  the  figure.  Then  a  line  marked  (a)  is  drawn 
anywhere  right  through  (1)  parallel  to  the  ray  a,  from  its  intersec- 
tion with  the  force  (1),  the  chord  (b)  is  drawn  parallel  to  the  ray  (6) 
in  Fig.  8b  meeting  (2);  through  this  latter  intersection  the  third 
chord  (c)  is  drawn  backward  parallel  to  its  reciprocal  the  ray  c. 
This  latter  is  the  closing  line  and  its  intersection  writh  the  initial 
line  (a),  gives  the  position  of  the  e.g.  of  the  two  forces. 


18  DAMS  AND  WEIRS 

A  vertical  line  through  this  center  of  pressure,  which  represents 
W,  i.e.,  Wi+Wz,  is  continued  on  to  the  profile  until  it  intersects 
the  inclined  force  PI  drawn  through  the  center  of  gravity  of  the 
water  pressure  area.  This  intersection  is  the  starting  point  of  R, 
drawn  parallel  to  its  reciprocal  on  the  force  polygon  8a.  This 
resultant  intersects  the  base  at  a  point  within  the  middle  third. 
R  is  the  resultant  "Reservoir  Full",  while  W,  the  resultant  of  the 
vertical  forces  in  the  masonry  wall,  is  the  resultant  "Reservoir 
Empty".  The  intersection  of  the  latter  is  almost  exactly  at  the  inner 
edge  of  the  middle  third — thus  the  condition  of  the  middle  third  is 
fulfilled.  The  question  of  induced  pressure  and  its  distribution  on 
the  base  will  be  considered  later. 

The  incidence  of  N,  the  vertical  component  "Reservoir  Full", 
on  the  base  is  naturally  not  identical  with  that  of  W,  the  resultant 
"Reservoir  Empty",  unless  the  back  of  the  wall  is  vertical.  The 
line  R  is  the  resultant  of  both  PI  and  W,  and  of  P  and  N.  If 
it  be  required  to  fix  the  position  of  N  on  the  profile,  a  horizontal 
line  should  be  drawn  through  the  intersection  of  PI  with  the  back 
of  the  wall.  This  will  represent  the  horizontal  component  of  the 
water  pressure  PI,  and  it  will  intersect  R,  produced  upward.  Then 
a  line  drawn  vertically  through  this  latter  point  will  represent  N, 
the  vertical  component  (Reservoir  Full).  The  position  of  N  is 
necessarily  outside  of  W,  consequently  if  N  is  made  to  fall  at  the  inner 
edge  of  the  middle  third  of  the  base,  W  must  fall  within  the  middle 
third.  This  fact  will  later  be  made  use  of  when  the  design  of  the 
lower  part  of  a  "high"  dam  comes  under  consideration. 

16.  Analytical  Method.  The  analytical  method  of  ascertain- 
ing the  positions  of  the  incidences  of  W  and  of  R  on  the  base,  which 
has  just  been  graphically  performed,  will  now  be  explained. 

The  first  step  is  to  find  the  positions  of  the  centers  of  gravity 
of  the  rectangle  and  trapezoid  of  which  the  profile  is  composed, 
relative  to  some  vertical  plane,  and  then  to  equate  the  sum  of  the 
moments  of  those  two  forces  about  any  fixed  point  on  the  base, 
with  the  moment  of  their  sum. 

The  most  convenient  point  in  most  cases  is  the  heel  of  the  base ; 
this  projects  a  distance  (y)  beyond  the  axis  of  the  dam,  which  axis 
is  a  vertical  line  passing  through  the  inner  edge  of  the  rectangular 
crest, 


DAMS  AND  WEIRS 


19 


As  the  areas  of  the  divisions,  whether  of  the  masonry  wall  or 
of  the  water-pressure  triangle,  are  generally  trapezoids,  the  follow- 
ing enumeration  of  various  formulas, 
whereby  the  position  of  the  e.g.  of  a 
trapezoid  may  be  found  either  with  regard 
to  a  horizontal  or  to  a  vertical  plane, 
will  be  found  of  practical  utility.  In  Fig. 
9,  if  the  depth  of  the  figure  between  the 
parallel  sides  be  termed  //,  and  that  of 
the  truncated  portion  of  the  triangle  of 
which  the  trapezoid  is  a  portion  be 
termed  d,  and  h  be  the  vertical  height 
of  the  e.g.  above  the  base,  then 


Fig.  9.     Diagram  Showing  Centers 

of  Gravity  of  Water  Pressure 

Trapezoids 


Thus,  in  Fig.  9,  #=13  and  d  =  Q>  feet,  then 


If  the  base  of  the  triangle  and  trapezoid  with  it  be  increased  or 
decreased  in  length,  the  value  of  h  will  not  be  thereby  affected, 
as  it  is  dependent  only  on  H  and  d,  which  values  are  not  altered. 
If,  however,  the  base  of  the  triangle  be  inclined,  as  shown  by  the 


Fig.  10.     Diagram  Illustrating  Height  of  c.  g.  Trapezoid  above  Base 

dotted  lines  in  Fig.  9,  the  center  of  gravity  of  the  trapezoid  will 
be  higher  than  before,  but  a  line  dfawn  parallel  to  the  inclined 
base  through  g,  the  c.  g.  will  always  intersect  the  upright  side  of 
the  trapezoid  at  the  same  point,  viz,  one  which  is  h  feet  distant 
vertically  above  the  horizontal  base. 


20 


DAMS  AND  WEIRS 


The  value  of  h  can  also  be  obtained  in  terms  of  a  and  6,  the  two 
parallel  sides  of  the  trapezoid,  and  is 

n       X. 

(6) 


For  example,  in  Fig.  10,  H  =  12,  a  =  10,  and  6  =  16,  then 

,     12/16+20\ 


If  the  horizontal  distance  of  the  e.g.  of  a  trapezoid  from  a  ver- 
tical plane  is  required,  as,  for  example,  that  of  the  trapezoid  in  Fig. 
8,  the  following  is  explanatory  of  the  working.  As  shown  in  Fig. 
11,  this  area  can  be  considered  as  divided  into  two  triangles,  the 
weight  of  each  of  which  is  equivalent  to  that  of  three  equal  weights 
placed  at  its  angles;  each  weight  can  thus  be  represented  by  one- 
third  of  the  area  of  the  triangle  in  question,  or  by  —  and  — ,  respec- 
tively, //  being  the  vertical  depth  of  the  trapezoid.  Let  y  be  the 

projection  of  the  lower  corner 
A  beyond  that  of  the  upper 
one  B.  Then  by  equating 
the  sum  of  the  moments  of 
the  corner  weights  about  the 
point  A  with  the  moment  of 
their  sum,  the  distance  (x~)  of 
the  e.g.  of  the  whole  trape- 
zoid from  A  will  be  obtained 
as  follows: 


(7) 


(7a) 


Fig.  11.     Method  of  Finding  Distance  of  Center 
of  Gravity  of  a  Trapezoid  from  Heel 


where  y  =  0,  the  formula  becomes 


a 
'a+b 


For  example,  in  Fig.  8,  a  or  de  =  5  feet,  6  =  33.3,  and  y  =  M,  whence 


DAMS  AND  WEIRS 


21 


The  similar  properties  of  a  triangle  with  a  horizontal  base,  as 
in  Fig.  12,  may  well  be  given  here  and  are  obtained  in  the  same  way 
by  taking  moments  about  A,  thus 

bh  ^         bh 
2-X.r  =  - 


(8) 


J+V 
3 

In  Fig.  12,  6  =  14  feet,  y  =  8  feet,  and  A  =  10  feet,  then 

14+8 
x  =  —  --  =  /J  feet 

O 

Reverting  to  Fig.  8,  the  position  of  the  incidence  of  W  on  the 
base  is  obtained  by  taking  moments  about  the  heel  g  of  the  base  as 
follows  :  Here  W  is  the  area  of  the  whole  profile,  equal,  as  we  have 
seen,  to  844  sq.  ft.  The  area  of  the  upper  component  (1)  is  40  sq. 
ft.  and  of  (2)  804. 

The  lever  arm  of  W  is  by  hypothesis  x, 
that  of  (1)  is  2.5+.84  =  3.34  feet,  that  of  (2) 
by  formula  (7)  has  already  been  shown  to  be 
1  1  .63  feet.  Hence,  as  the  moment  of  the  whole 
is  equal  to  the  sum  of  the  moments  of  the 
parts,  the  equation  will  become 

844z  =  40  X  3.34+804  X  1  1  .63  =  9484.1  _  Fig-  .12.   Method  of 

Finding  Center  of 
Gravity  of  a  Triangular 
Profile 

x  =  11.  23  feet 
This  fixes  the  position  of  the  incidence  of  W  relative  to  the  heel. 

7  QQ    O 

The  position  of  the  inner  third  point  is  —  ,  or  '-^—  from  the  heel. 

3          o 

The  incidence  of  W  is  therefore  11.23-11.  10  =  .  13  foot  within  the 
middle  third,  which  complies  with  the  stipulated  proviso. 

The  next  step  is  to  find  the  position  of  R  relative  to  the  heel  of 
the  base.  As  in  graphical  methods,  only  horizontal  and  vertical 
forces  are  considered;  the  water-pressure  area  is  split  into  two  parts, 


H2 

one,  P  the  horizontal  component,  the  value  of  which  is  -r— 


,  or  555 


feet,  and  w3  the  reduced  area  of  water  overlying  the  rear  projection 
of  the  back.    The  latter  area  is  a  trapezoid  of  which  the  upper  side 


22  DAMS  AND  WEIRS 

(a)  is  8  feet  long  and  6,  the  lower  side,  50  feet,  the  depth  being  .84 
foot,  hence  the  distance  of  its  c.  g.  inside  the  heel  of  the  base  will  be 

by  formula  (6),  ^  (50+16)  =  .32  foot.     Its  actual  area  is  ^±™X 

o  Do  2 

.84  =  24.4  feet;  this  has  to  be  divided  by  p  or  2J  to  reduce  it  to  a 
masonry  base.  The  reduced  area  will  then  be  10.8  square  feet, 
nearly.  The  distance  of  the  incidence  of  W  from  the  heel  of  the 
base  has  already  been  determined  to  be  11.23  ft.  and  that  of  w9 
being  .32  ft.,  the  distance  of  the  c.  g.  of  the  latter  from  W  will  be 
11.23  —  .32  =  10.9,  nearly.  If  the  distance  between  the  incidences  of 
W  and  R  be  termed  x,  the  equation  of  moments  about  the  incidence 
of  R,  will  stand  thus: 


or 


.e. 

9 

=  10.7  ft.,  newly 


R  is  therefore  10.7+11.23  =  21.93  ft.  distant  from  the  heel.  The 
§  point  being  22.2  ft.  from  the  same  point,  R  falls  .3  ft.  (nearly) 
within  the  middle  third.  This  shows  that  a  small  reduction  in  th( 
area  of  the  profile  could  be  effected. 

17.    Vertical  Component.      If  the  position  of  N,  the  vertical 

component   of   R  and  PI,  is  required,  as  is   sometimes  the  case, 

it    is    obtained    by    the   equation    NXx  =  (WX  11.23)  +  (w8X.  32), 

x  being  the  distance  from  the  heel  of  the   base.     Or  in  figures, 

854.8s  =  (844  X  1  1  .23)  +  (10.8  X  .32) 

z  =  ll.l  feet 

The  incidence  of  N  is,  therefore,  in  this  case,  exactly  on  the  limit 
the  middle  third.  This  of  course  does  not  affect  the  condition  of 
middle  third,  which  refers  to  the  resultant  W  (R.E.)  not  to  the  com- 
ponent N  (R.F.)  but,  as  will  be  seen  later,  when  the  lower  part  of  a 
high  dam  comes  to  be  designed,  one  condition  commonly  imposed 
is,  that  the  vertical  component  N  must  fall  at  the  inner  edge  of  the 
middle  third,  in  which  case  W  will  necessarily  fall  inside  thereof. 


DAMS  AND  WEIRS  23 

It  may  here  be  noted  that  the  space  between  the  location  of  N  and 

P  II 

R,  which  will  be  designated  (/),  is  —  —  because  if  moments  are  taken 

o  J\ 

PII  P  II 

about  the  incidence  of  R,  then  Nf=—  —  ;  therefore  /=—r:r.    The 

O  O  JL\ 

actual  value  of  W  in  tons  of  2000  pounds  will  be  the  superficial  area, 
or  844  square  feet  multiplied  by  the  eliminated  unit  weight,  i.e.,  by 

844  X  9  1 

trp,  viz,  -7-^  -  =  59.3  tons,  as  w=—  ton.     That  of  the  inclined 
32X4  o2 

force  R,  is  obtained  from  the  triangle  of  forces  PNR  in  which  R, 


being  the  hypothenuse  =  ^J  N'2+P2.  Here  N  =  855  square  feet, 
equivalent  to  60  tons,  nearly,  and  P  =  555  feet,  equivalent  to  39 

tons,  whence  fl  =  V602+392  =  71.5  tons. 

18.  Pressure  Distribution.  In  the  design  of  the  section  of  a 
dam,  pier,  or  retaining  wall,  the  distribution  of  pressure  on  a  plane 
in  the  section  and  the  relations  existing  between  maximum  unit 
stress,  symbolized  by  (s),  and  mean  or  average  unit  stress  (si)  will 
now  be  considered.  The  mean  unit  stress  on  any  plane  is  that 
which  acts  at  its  center  point  and  is  in  amount  the  resultant  stress 
acting  on  the  plane  (the  incidence  of  which  may  be  at  any  point) 
divided  by  the  wTidth  of  the  lamina  acted  on.  Thus  in  Figs.  3  or  8 
take  the  resultant  W.  This  acts  on  the  horizontal  base  and  its 

mean  unit  stress  Si  will  be  —  .     In  the  same  way,  with  regard  to  N,  the 

vertical  component  of  R  the  mean  unit  stress  produced  by  it  on  the 

N 

horizontal  base  will  be  —  .     The  maximum  unit  stress  occurs  at 

b 

that  extremity  of  the  base  nearest  to  the  force  in  question  which  is 
R.  Thus  the  maximum  unit  stress  due  to  W  is  at  the  heel  while 
that  due  to  a  combination  of  P  and  N  acting  at  the  incidence  of  R 
is  at  the  toe  of  the  base  b.  It  is  evident  that  the  nearer  the  inci- 
dence of  the  center  of  pressure  is  to  the  center  point  the  less  is  the 
maximum  stress  developed  at  the  outer  edge  of  the  section,  until 
the  center  of  pressure  is  actually  situated  at  the  center  point  itself. 
The  maximum  pressure  at  the  outer  part  of  the  section  then  equals 
the  average  and  is  thus  at  a  minimum  value.  The  relation  between 
maximum  and  mean  unit  stress  or  reaction  is  expressed  in  the  fol- 


24 


DAMS  AND  WEIRS 


lowing  formula  in  which  it  is  assumed  that  any  tension  at  the  heel 
can  be  cared  for  by  the  adhesion  of  the  cementing  material  or  of 
reinforcement  anchored  down : 

X  /?,JV 

(9) 


or,  letting  m  equal  the  expression  in  brackets, 

•  -IM,  <9a) 

In  formula  (9a),  q  is  the  distance  between  the  center  point  of  the 
base  and  the  center  of  pressure  or  incidence  of  whatever  resultant 

pressure   is   under  considera- 
tion, and  Si  is  the  mean  stress, 
the  resultant  pressure  di- 
ps?*   vided  by  the  base. 

In  Fig.  8  as  explained  in 
section  16,  the  incidence  of  R, 
i.e.,  the  center  of  pressure 
(R.F.),  falls  .3  ft.  within  the 
middle  third  of  the  base,  con- 
sequently the  value  of  q  will  be 

A_.3  =  ?|3_.3=5.25ft.,and 


in  formula    (9a)    m  =  1  +  -~ 
=  1+1^1  =  1.95.     The  maxi- 

oo.o 
mum     reaction    at    the    toe 

always  designated  by  s  =  — — 


Fig.  13.     Diagram  Showing  Pressure  Distribution 

on  a  Dam  with  Reservoir  Empty  and 

Reservoir  Full 


the  heel,  m  =  1  -  .95  =  .05,  and  s2  = 


1.95X60 


=  3.51  tons  per  sq. 


33.3 
ft.     For  the  reaction  (R.F.)  at 

=  .09  tons.     The  distribu- 


OO.3 


tion  of  pressure  due  to  the  vertical  component  of  #  is  shown  hatched 
in  Fig.  8  as  well  as  in  Fig.  13. 

From  formula  (9)  the  facts  already  stated  are  patent.     When 
the  incidence  of  the  resultant  force  is  at  the  center  of  the  base, 


DAMS  AND  WEIRS  25 

q  =  Q,  consequently  m  =  l  and  s  =  Si,  that  is,  the  maximum  is  equal 

to  the  mean ;  when  at  one  of  the  third  points,  q  =  — ,  m  =  2,  and  s  =  2s\ ; 

6 

IF 
when  at  the  toe,  m  =  4,  and  s  =  4#i,  or  4—. 

If  the  material  in  the  dam  is  incapable  of  caring  for  tensile 
strain,  the  maximum  vertical  compression,  or  s,  obtained  by  formula 
(9)  will  not  apply.  Formula  (24),  section  86,  should  be  used  when- 
ever R  falls  outside  the  middle  third. 

In  designing  sections  it  is  often  necessary  to  maneuver  the 
incidence  of  the  resultant  stress  to  a  point  as  close  as  possible  to  the 
center  of  the  base  in  order  to  reduce  the  maximum  stress  to  the  least 
possible  value,  which  is  that  of  the  mean  stress.  The  condition  of  the 
middle  third,  insures  that  the  maximum  stress  cannot  exceed  twice 
the  mean,  and  may  be  less,  and  besides  insures  the  absence  of  tensile 
stress  at  the  base. 

19.  Graphical  Method  for  Distribution  of  Pressure.  The 
graphical  method  of  ascertaining  the  distribution  of  pressure  on  the 
base  of  a  masonry  wall,  which  has  already  been  dealt  with  analyti- 
cally, is  exhibited  in  Fig.  13,  which  is  a  reproduction  of  the  base  of 
Fig.  8.  The  procedure  is  as  follows:  Two  semicircles  are  struck 
on  the  base  line,  having  their  centers  at  the  third  division  points 

and  their  radii  equal  to  — .     From  the  point  marked  e,  that  of  the 

o 

incidence  of  R,  the  line  eg  is  drawn  to  g,  the  point  of  intersection  of 
the  two  semicircles.  Again  from  g  a  line  gn  is  set  off  at  right 
angles  to  eg  cutting  the  base  or  its  continuation  at  a  point  n.  This 
point  is  termed  the  antipole  of  e,  or  the  neutral  point  at  which  pres- 
sure is  nil  in  either  sense — compressive  or  tensile.  Below  and 
clear  of  the  profile  a  projection  of  the  base  is  now  made,  and  from 
g  a  perpendicular  is  let  fall,  cutting  the  new  base  in  g'  while,  if  the 
line  be  continued  upward,  it  will  intersect  the  base  at  K.  This 
latter  point  will,  by  the  construction,  be  the  center  point  of  the  base. 
The  line  Kg  is  continued  through  g'  to  h* ',  g'hf  being  made  equal  to 
the  mean  unit  pressure,  =1.8  tons.  A  perpendicular  is  let  fall 
from  n  cutting  the  new  base  line  at  ni;  the  points  n\  and  h'  are  then 
joined  and  the  line  continued  until  it  intersects  another  perpendicu- 
lar let  fall  from  the  toe  of  the  base.  A  third  perpendicular  is  drawn 


26 


DAMS  AND  WEIRS 


from  the  heel  of  the  base,  cutting  off  a  corner  of  the  triangle.  The 
hatched  trapezoid  enclosed  between  the  last  two  lines  represents  the 
distribution  of  pressure  on  the  base.  The  maximum  stress  will 
scale  close  upon  3.51  and  the  minimum  .09  tons.  If  W  be  considered, 

W    59.3 

s  =  —  =  -^'—=l.78  tons,  the  maximum  stress  at  the  heel   will  be 
b      33.3 

3.52  and  the  minimum  .04,  at  the  toe. 

20.     Examples  to  Illustrate  Pressure  Distribution.    In  Fig.  14 
is  illustrated  the  distribution  of  pressure  on  the  base,  due  to  the 

incidence  of  R,  first,  at  the  toe  of 
the  base,  second,  at  the  two-third 
point,  third,  at  the  center,  and 
fourth,  at  an  intermediate  position. 
In  the  first  case  (7?i),  it  will  be 
seen  that  the  neutral  point  n\  falls 
at  the  first  third  point.  Thus  two- 
thirds  of  the  base  is  in  compression 
and  one-third  in  tension,  the  maxi- 
mum in  either  case  being  propor- 
tional to  the  relative  distance  of  the 
neutral  point  from  the  toe  and  heel 
of  the  base,  the  compression  at  the 
toe  being  four  times,  while  the  ten- 
sion at  the  heel  is  twice  the  mean 
stress.  In  the  second  case  (Rz) 
intersects  at  the  two-third  point, 
and  the  consequent  position  of  n  is 
exactly  at  the  heel.  The  whole  base 
is  thus  in  compression,  and  the  max- 
imum is  double  the  mean.  In  the 
third  case  (Rs),  the  line  gn  is 
drawn  at  right  angles  to  fg.  The 
latter  is  vertical  and  gn  will  conse- 
quently be  horizontal.  The  distance  to  n  is  thus  infinite  and  the 
area  of  pressure  becomes  a  rectangle  with  a  uniform  unit  stress  s. 
In  the  fourth  case  (#4),  the  neutral  point  lies  well  outside  the  profile, 
consequently  the  whole  is  in  compression,  the  condition  approximating 
to  that  of  R3. 


Fig.   14.     Pressure  Distribution  on 

Base  of  Dam  under  Various 

Conditions 


DAMS  AND  WEIRS  27 

21.  Maximum     Pressure     Limit.     The     maximum    pressure 
increases  with  the  depth  of  the  profile  until  a  level  is  reached  where 
the  limit  stress  or  highest  admissible  stress  is  arrived  at.     Down  to 
this  level  the  design  of  the  section  of  a  dam,  as  already  shown,  con- 
sists simply  in  a  slight  modification  of  the  pentagonal  profile  with 
a  vertical  back,  the  base  width  varying  between  that  of  the  ele- 

mentary profile  or  -—  ,  or  its  reduced  value  given  in  formula  (3). 

VP 

Beyond  this  limiting  depth,  which  is  the  base  of  the  so-termed 
"low"  dam,  the  pentagonal  profile  will  have  to  be  departed  from 
and  the  base  widened  out  on  both  sides. 

22.  Formulas  for  Maximum  Stress.    The  maximum  unit  stress 
in  the  interior  of  a  dam  is  not  identical  with  (s),  the  maximum  vertical 
unit  reaction  at  the  base,  but  is  a  function  of  Si.     In  Fig.  8,  a  repre- 
sentative triangle  of  forces  is  shown  composed  of   N  the  vertical 
force  (R.F.),  P  the  horizontal  water  pressure,  and  R  the  resultant 
of  N  and  P;  therefore  R  =  V  #2+P2  =  also  N  sec  6.     If  the  back 
were  vertical,    N  and  W  would  coincide  and  then  R  =  ^J 


Various  views  have  been  current  regarding  the  maximum  internal 
stress  in  a  dam.  The  hitherto  mos.t  prevalent  theory  is  based  on 
the  assumption,  see  Fig.  8,  that  the  maximum  unit  stress 


Another  theory  which  still  finds  acceptance  in  Europe  and  in  the 
East  assumes  that  the  maximum  stress  is  developed  on  a  plane 
normal  to  the  direction  of  the  resultant  forces  as  illustrated  by  the 
stress  lines  on  the  base  of  Fig.  8.  According  to  this,  the  mean 

7?  7? 

stress  due  to  R  would  not  be  --  but  —  ,  and  the  maximum  stress  will 

o  61 

,      mR      _      R      RsecO 

be  —  —  .     But  T-  =  —       -  and  R=  N  sec  6,  consequently  the  maxi- 
61  61          6 

mum  unit  stress  would  be 

C  =  ?l^sec20  (10b) 

0 

Recent  experiments  on  models  have  resulted  in  the  formula  for 
maximum  internal  unit  stress  being  recast  on  an  entirely  different 
principle  from  the  preceding.  The  forces  in  action  are  the  maxi- 


28  DAMS  AND  WEIRS 

mum  vertical  unit  force  or  reaction  s  combined  with  a  horizontal 

p 
shearing  unit  stress  ss=—.     The  shearing  force  is  the  horizontal 

o 

water  pressure,  or  -  —  -  symbolized  by  P,  which  is  assumed  to  be 
2p 

equally  resisted  by  each  unit  in  the  base  of  the  dam;  the  unit  shear- 

p 
ing  stress  will  thus  be  —  .    These  forces  being  at  right  angles  to 

6 

each  other,  the  status  is  that  of  a  bar  or  column  subject  to  compres- 
sion in  the  direction  of  its  length  and  also  to  a  shear  normal  to  its 
length.  The  combination  of  shear  with  compression  produces  an 
increased  compressive  stress,  and  also  a  tension  in  the  material. 
The  formula  recently  adopted  for  maximum  unit  compression  is 
as  follows  : 

c  =  i*+Vf^+v  (10) 

In  this  s  =  msi,  =  —  —  .     As  before  ss=—  ,   substituting   we    have 
6  b 


_mN  ,     l(mN)2  ,  P2_m]V+V(mA02+4P2       in  , 
=  26  +\    4&2     U2~  26 

When  m  =  2,  as  is  the  case  when  the  incidence  of  R  is  exactly  at  the 
outer  boundary  of  the  middle  third 


=-=-  (1+  sec  0)  (102) 


b  b 

23.    Application  of  All  Three  Formulas  ,to  Elementary  Profile. 

In  the  case  of  elementary  triangular  profile  which  has  a  vertical 

back,  N  =  W  and  sec  6=    P~~     (section  7,  page  5)  and  m  =  2;  then 
formula  (102)  becomes 


Now 


c  = 


(ii) 


DAMS  AND  WEIRS  29 


Example. 


Let  H  in  elementary  triangle  =  150  feet,   p  =  2.4,  wp  =  —  ton. 


When,   according   to    (11),   c=        7      (1  +  1. 187)  =  12.3   tons   per 

L  X40 

square  foot. 

Taking  up  formula  (lOa) 

mW  sec  e     2W  sec  6 

C==—T        — 

,         W     Hwp 
as  above  -7-  =  — — 

o         2 


c=  Hwp\- —  =  #wVpVp+i  (Ha) 

p 


Example  with  conditions  as  before 
150X1  XL55V3.4 


32 


=  7.26X1.84  =  13.3  tons 


2W  sec2  6 
With  formula  (lOb),  c  = - ,  or  in  terms  of  H, 


c=  Hwp  I  ^  j  =  Hw(p+l)  (lib) 

Therefore,  with  values  as  above, 

150X1X3.4 
c  = =  15.9  tons 

From  the  above  it  is  evident  that  formulas  (lOb)  and  (lib)  give 
a  very  high  value  to  c.  Tested  by  this  formula,  high  American 
dams  appear  to  have  maximum  compressive  unit  stresses  equal  to 
20  tons  per  square  foot,  whereas  the  actual  value  according  to 
formula  (10)  is  more  like  14  tons.  However,  the  stresses  in  the 
Assuan  dam,  the  Periyar,  and  other  Indian  dams,  as  also  French 
dams  have  been  worked  out  from  formula  (lOb)  which  is  still  in  use. 

24.  Limiting  Height  by  Three  Formulas.  The  limiting  height 
(Hi)  of  the  elementary  triangular  profile  forms  a  close  guide  to 
that  obtaining  in  any  trapezoidal  section,  consequently  a  formula  will 
be  given  for  each  of  the  three  cases  in  connection  with  formulas  (10), 
( 1  Oa) ,  and  ( 1  Ob) .  Referring  to  case  ( 1 0) ,  we  have  from  formula  (11) 

Hwp, 


30  DAMS  AND  WEIRS 

9, 


Whence  ///,  the  limiting  height  = 


wp 


Example. 

With  c=16  tons  and  p  =  2.4,  HI,  the  limit  height  of  the  ele- 

2X16X33     1024 
mentary  profile  will  be  =  =19o  feet. 


Referring  to  case  (lOa),  we  have  from  formula  (11  a) 
c=Hw  VpVp+1 

Hi=- 


U'VpVp  +  1 
Example. 

16X32 

With  data  as  above  //  =  —  —  =  180  feet,  nearly. 

1.55X1.84 

Referring  to  case  (lOb),  we  have  from  formula  (lib) 
c=Hw(p+l) 


Example. 

With  same  data  Hl=  =        =  l^  feet. 

o.4         o.4 

Thus  the  new  formula  (10)  gives  much  the  same  results  as  that 
formerly  in  general  use  in  the  United  States  (lOa),  while  in  the 
more  conservative  formula  (lOb)  the  difference  is  marked. 

25.  Internal  Shear  and  Tension.  We  have  seen  that  the 
combination  of  compressive  and  shearing  stresses  in  a  dam  (R.F.) 
produces  an  increased  unit  compression.  It  further  develops  an 
increase  in  the  shearing  stress  and  also  a  tensile  stress.  The  three 
formulas  are  given  below. 
Compression  as  before 


(10) 

Tension 


DAMS  AND  WEIRS  31 

Shear 

— h^-2  or 


4 


The  tensile  and  shearing  stresses  are  not  of  sufficient  moment  to  require 
any  special  provision  in  the  case  of  a  gravity  dam.  The  tension  is 
greatest  at  the  heel,  diminishing  toward  the  toe.  This  fact  suggests 
that  a  projection  of  the  heel  backward  would  be  of  advantage.  The 
direction  (a)  of  c  to  the  vertical  is  not  that  of  R  but  is  as  follows: 

9<?        9P     m  V       9P 

T0            Z.O,          ^1  II I  ^  LiL 

an  2a= =  -r+- T~  =  ~ ^f 

sob       mN 

p 
when m  =  2,  tan 2a  =  — .     In  Fig.  8,  P  =  555  and  N  =  855.    .' .  tan  2a  = 

N 

—  =  .649  whence  2a  =  33°  00'  and   (a) -16°  30'.     The  inclination 

855 

of  R  to  the  vertical,  or  B,  is  33°  50',  i.e.,  twice  as  large  as  that 
of  c.  The  direction  of  t  is  at  right  angles  to  that  of  c,  while  that  of 
Sh  the  shear,  lies  at  45°  from  the  directions  of  either  c  or  t. 

26.  Security  against  Failure  by  Sliding  or  Shear.    Security 
against  failure  by  sliding  depends  on  the  inclination  of  W  to  P,  i.e., 
on  the  angle  B  between  W  and  R.    Thus  tan  B  should  be  less  than  the 
angle  of  friction  of  masonry  on  masonry,  or  less  than  .7.     This  is  the 
same  as  stating  that  the  relation  of  W  and  P  must  be  such  that  B 
shall  not  be  greater ( than  35°,  or  that  the  complement  of  B  be  not  less 
than  55°.     The  adoption  of  the  middle  third  proviso  generally  insures 
this.     With  regard  to  sliding  on  the  base,  this  can  be  further  pro- 
vided against  by  indentations  in  the  base  line  or  constructing  it 
inclined  upward  from  heel  to  toe. 

27.  Influence  Lines.     It  is  sometimes  desirable  for  the  purpose 
of  demonstrating  the  correctness  of  a  profile  for  tentative  design, 
to  trace  the  line  of  pressures  corresponding  to  the  two  conditions 
of  reservoir  full  and  empty,  through  the  profile  of  a  dam.     This  is 
far  better  effected  by  the  use  of  graphic  statics. 

There  are  two  different  systems  of  graphic  construction  that 
give  identical  results,  which  will  now  be  explained  and  illustrated. 

The  first  method,  which  is  most  commonly  adopted,  is  exhibited 
in  Fig.  15,  which  is  the  profile  of  a  100-foot  high  dam  with  specific 
gravity  2J.  It  thus  lies  within  the  limiting  depth,  which  for  the 
elementary  profile  would  be  190  feet. 


32 


DAMS  AND  WEIRS 


The  profile  is  pentagonal,  with  a  vertical  back,  and  has  the  full!  ri 

base  width  of  the  elementary  profile,  viz,  •— =•  which  in  this  case  isf 

Vp 

?X  100  =  66.7  feet.    The  crest  k  is  Vlf=10  feet  wide.     The  water- 

o 

pressure  triangle  has  a  base  of  — .     The  profile,  as  well  as  the  water- 

P 

pressure  triangle,  is  divided  into  five  equal  laminas,  numbered  1 


ft*?? 


^0,Z5  Tons 

Fig.  15.     Graphical  Construction  for  Tracing  Lines  of  Pressure  on  Dam  of  Pentagonal  Profile 

to  5  in  one  case  and  ]/  to  5'  in  the  other.     The  depth  of  each  lamina, 

TJ 

which  is  —  is,  therefore,  a  common  factor  and  can  be  eliminated  as 
5 

well  as  the  item  of  unit  weight,  viz,  wp.  The  half,  widths  of  all 
these  laminas  will  then  correctly  represent  their  areas  and  also  their 
weights,  reduced  to  one  denomination,  that  of  the  masonry.  In 
Fig.  15a  a  force  polygon  is  formed.  In  the  vertical  load  line  the 
several  half  widths  of  the  laminas  1  to  5  are  first  set  off,  and  at 


DAMS  AND  WEIRS  33 

i  right  angles  to  it  the  force  line  of  water  pressure  is  similarly  set 
out  with  the  half  widths  of  the  areas  1'  to  5'.     Then  the  resultant 
lines  of  the  combination  of  1  with  I',  1,  2,  with  V  2'  and  so  on  marked 
R!  to  7?5  are  drawn.     This  completes  the  force  polygon.     The  next 
step  is  to  find  the  combinations  of  the  vertical  forces  on  the  profile, 
viz,  that  of  1  and  2,  1,  2,  and  3,  etc.    This,  as  usual,  is  effected  by 
•  constructing  a  force  and  ray  polygon,  utilizing  the  load  line  in 
Fig.  15a  for  the  purpose.     Then  the  centers  of  gravity  of  the  several 
individual  areas  1  to  5  are  found  by  the  graphical  process  described 
in  section  15,  and  verticals  drawn  through  these  points  are  projected 
below  the  profile.     On  these  parallel  force  lines  1  to  5,  the  funicular 
>  polygon  Fig.  15b  is  constructed,  its  chords  being  parallel  to  their 
!  reciprocal  rays  in  Fig.  15a.    The  intersection  of  the  closing  lines 
j  of  the  funicular  gives  the  position  of  the  centroid  of  the  five  forces 
I  ngaged.     By  producing  each  chord  or  intercept  backward  until  it 
i  intersects  the  initial  line,  a  series  of  fresh  points  are  obtained  which 
i  denote  the  centers  of  gravity  of  the  combinations  of  1  and  2;  1,  2,  and 
i  3,  and  so  on.    Verticals  through  these  are  next  drawn  up  on  the  profile 
i  so  as  to  intersect  the  several  bases  of  the  corresponding  combinations, 
thus  1,  2,  and  3  will  intersect  the  base  of  lamina  3;  and  1,  2,  3,  and 
4  will  intersect  that  of  lamina  4;  and  so  on.     These  intersections 
are  so  many  points  on  the  line  of  pressure  (R.E.).     The  next  step 
is  to  draw  the  horizontal  forces,  i.e.,  their  combinations  on  to  the 
profile.     The  process  of  finding  the  centers  of  gravity  of  these  areas 
is  rendered  easy  by  the  fact  that  the  combinations  are  all  triangles, 
not  trapezoids,  consequently  the  center  of  gravity  of  each  is  at  J 
its  height  from  the  base.    Thus  the  center  of  gravity  of  the  com- 
bination l/+2'4-3'  is  at  J  the  height  measured  from  the  base  of 
3'  to  the  apex,  in  the  same  way  for  any  other  combination,  that  of 
1',  2',  3',  4',  5',  being  at  J  the  total  height  of  the  profile.    The  back 
being  vertical,  the  direction  of  all  the  combined  forces  will  be  hori- 
zontal, and  the  lines  are  drawn  through,  as  shown  in  the  figure, 
to    intersect   the    corresponding    combinations   of   vertical   forces. 
Thus  1'  intersects  1,  I'  2'  intersects  1  2,  and  so  on.    From  these 
several  intersections  the  resultant  lines  RI,  Rz,  to  7?5  are  now  drawn 
:lown  to  the  base  of  the  combination  to  which  they  belong,  these 
last  intersections  giving  the  incidence  of   R\,  Rz,  etc.,  and  are  so 
many  points  on  the  line  of  pressures  (R.F.).    The  process  is  simple 


34  DAMS  AND  WEIRS 

and  takes  as  long  to  describe  as  to  perform,  and  it  has  this  advan- 
tage, that  each  combination  of  forces  is  independent  of  the 
rest,  and  consequently  errors  are  not  perpetuated.  This  system 
can  also  be  used  where  the  back  of  the  profile  has  one  or 
several  inclinations  to  the  vertical,  explanation  of  which  will  be 
given  later. 

28.    Actual  Pressures  in  Figures.     In  the  whole  process  above 
described,  it  is  noticeable  that  not  a  single  figure  or  arithmetical 
calculation  is  required.     If  the  actual  maximum  unit  stress  due  to' 
R  or  to  W  is  required  to  be  known,  the  following  is  the  procedure. 
In  Fig.  15a,  N  scales  174,  to  reduce  this  to  tons  it  has  to  be  mul- 

tiplied by  all  the  eliminated  factors,  which  are  —  =  20    and    wp  = 

o 

9X1    4.1,  *•     AT    174X20X9 
—  -.that  is,  2V  =    -- 


Assuming  the   incidence   of  R  exactly  at  the  third  division, 
the  value  of  q  is  —  and  that  of  m  is  2;  P  also  scales    112,  its 

112x20X9 
value  is  therefore  —  -  —  —  -  —  =  157  tons.     Applying  formula  (102), 


A7  .  244+V2442+1572     534  - 

c=N  + —  -=__  =  8   tons   per   sq.    ft., 

b  60.7  66.7 

roughly.  As  8  tons  is  obviously  well  below  the  limiting  stress,  for 
which  a  value  of  16  tons  would  be  more  appropriate,  this  estimation 
is  practically  unnecessary  but  is  given  here  as  an  example. 

29.  Analytical  Method.  The  analytical  method  of  calculation 
will  now  be  worked  out  for  the  base  of  the  profile  only.  First  the 
position  of  W,  the  resultant  vertical  forces  (R.E.)  relative  to  the 
heel  of  the  base  will  be  calculated  and  next  that  of  R.  The  back  of 
the  profile  being  in  one  line  and  vertical  the  whole  area  can  be  con- 
veniently divided  into  two  right-angled  triangles,  if  the  thick- 
ening of  the  curvature  at  the  neck  be  ignored.  As  the  fore  slope 
has  an  inclination  of  1 :  Vp  the  vertical  side  of  the  upper  triangle 

,   /-  .  A-2Vp     100X1.5 

(1)  is  k\p  in    length;    its    area   will  then  be  — - —        — - —    =75 

sq.  feet.     The  distance  of  its  c.  g.  from  the  heel  of  the  base,  which 

20 

in  this  case  corresponds  with  the  axis  of  the  dam,  is  —  feet  =  6§  feet. 
•  «j 


DAMS  AND  WEIRS  35 

The  moment  will  then  be  —  -  —  =  500.     With  regard  to  the  lower 

o 

7T2  9 

triangle,  its  area  is  —  7=  =  5000X-  =  3333.3  sq.  feet.     The  length  of 

2Vp  3 

its  lever  arm  is  one-third  of  its  base,  or  22.2  feet.  The  moment 
about  the  axis  will  then  be  3333.3X22.2  =  74,000.  The  moment  of 
the  whole  is  equal  to  the  sum  of  the  moments  of  the  parts.  The 
area  of  the  whole  is  75+3333  =  3408.  Let  x  be  the  required  distance 
of  the  incidence  of  W  from  the  heel,  then 

xX  3408  =  74,500 


The  inner  edge  of  the  middle  third  is  —  or  22.2  feet  distant  from  the 

heel;  the  exact  incidence  of  W  is,  therefore,  .3  foot  outside  the  middle 
third,  a  practically  negligible  amount.  With  regard  to  the  position 
of  R  the  distance  (/)  between  the  incidence  of  R  and  that  of  W  is 

PH  100X444 

;  in  this  P  the  water-pressure  area=-  —  =  2220.     .'./= 


999f)  -y  1  AQ 

^  =  21.7  feet.     The  total  distance  of  R  from  the  heel  will 
o  X  o4(Jo 

then  be  21.7+21.9  =  43.6  feet;  the  outer  edge  of  the  middle  third 
is  44.4  feet  distant  from  the  heel,  consequently  the  incidence  of  R 

is  44.4  -43.6  =  .8  foot  within  the  middle  third,  then  q  =  —  -.8=— 

6  6 

-.8  =  10.3   ft.,   and  m  =  (l  +^\  =  1  +  .93  =  1.93.     At  this  stage  it 

will  be  convenient  to  convert  the  areas  into  tons  by  multiplying 

2  25 
them  by  pw,  or  -^—  .    Then  N  and  W  become  239.6,  and  P  becomes 

156.3  tons.     Formula  (10)  will  also  be  used  on  account  of  the  high 
figures;  then 


„  mW    239.6X1.93  P     156.3 

Here   s=—  -=  --  —  —  --  =  6.93   tons,    and   5s=  —  =  -—  -  =  2.34 
o  66.7  b       66.7 


tons,  therefore,  c  =        +-  +  (2.34)2  =  3.46+l7.48  =  7.64  tons. 


36 


DAMS  AND  WEIRS 


T-  ^  u    i         i     69      AT  239.6X0.07 

For  «2,  or  the  compression  at  the  heel,  m  =  1  — —  =  .07.  52  =  - 

o  DO. 7 

=  .251  ton.     The  area  of  base  pressure  is  accordingly  drawn  on  Fig.  15. 
If  W    (R.E.)    be   considered,    g  =— +.3  =  11.42,    and   m  =  l  + 

6X11.42  mW    239.6X2.03     .  f 

-=2.03;  therefore,  s  =  —:—  =  —  -=7.30  tons.     The 

66.7  0  66.7 

base  pressure  is  therefore  greater  with  (R.E.)  than  with  (R.F.); 


f.2'    3'         4'  5' 


(a) 


Fig.  16.     Diagram  Showing  Haessler's  Method  for  Locating  Lines  of  Pressure  on  a  Dam 

there  is  also  a  slight  tension  at  the  toe  of  .11  ton,  a  negligible 
quantity.    This  pressure  area  is  shown  on  Fig.  16. 

30.  Haessler's  Method.  A  second  method  of  drawing  the 
line  of  pressures  which  is  termed  "Haessler's"  is  exhibited  in  Fig. 
16,  the  same  profile  being  used  as  in  the  last  example.  In  this 
system,  widen  is  very  suitable  for  a  curved  back,  or  one  composed 
of  several  inclined  surfaces,  the  forces  are  not  treated  as  independent 
entities  as  before,  but  the  process  of  combination  is  continuous 
from  the  beginning.  They  can  readily  be  followed  on  the  force 
polygon,  Fig.  16a  and  are  I'  with  1  producing  ft;  ft  with  2',  i.e., 
1',  1,  2r,  the  last  resultant  being  the  dotted  reverse  line.  This  last 
is  then  combined  with  2  producing  ft,  and  so  on. 


DAMS  AND  WEIRS 


37 


The  reciprocals  on  the  profile  are  drawn  as  follows:  First  the 
c.g.'s  of  all  the  laminas  1,  2,  3,  etc.,  1',  2',  3',  etc.,  are  obtained  by 
graphical  process.  Next  the  water-pressure  lines,  which  in  this  case 
are  horizontal,  are  drawn  through  the  profile.  Force  line  (!')  inter- 
sects the  vertical  (1),  whence  Ri  is  drawn  parallel  to  its  reciprocal 
in  Fig.  16a  through  the  base  of  lamina  (1),  until  it  reaches  the  hori- 
zontal force  line  (20.  Its  intersection  with  the  base  of  (1)  is  a  point 
in  the  line  of  pressure  (R.F.).  Again  from  the  intersection  of  Ri 
with  (20,  a  line  is  drawn  backward  parallel  to  its  dotted  reciprocal 
line  in  Fig.  16a  until  it  meets  with  the  second  vertical  force  (2). 
From  this  point  R2  is  then  drawn  downward  to  its  intersection  with 
the  horizontal  force  (30,  its  intersection  with  the  base  of  lamina 
(2)  giving  another  point  on  the  line  of  pressures.  This  process  is 
repeated  until  the  intersection  of  R$  with  the  final  base  completes 
the  operation  for  (R.F.).  It  is  evi- 
dent that  Rs  as  well  as  all  the  other 
resultants  are  parallel  to  the  cor- 
responding ones  in  Fig.  15,  the  same 
result  being  arrived  at  by  different 
graphical  processes. 

31.  Stepped  Polygon.    Fig. 
16b  is  a  representation  of  the  so- 
called    "stepped"    polygon,    which 
is  also  often  employed;  the  form 

differs,  but  the  principle  is  identical  with  that  already  described. 
Inspection  of  the  figure  will  show  that  all  the  resultant  lines  are 
drawn  radiating  to  one  common  center  or  nucleus  (0). 

The  process  of  finding  the  incidence  of  W  on  the  bases  of  the 
several  lamina  is  identical  with  that  already  described  with  regard 
to  Fig.  15,  viz,  the  same  combination  of  1+2,  1+2+3,  and  so  on, 
are  formed  in  the  funicular  16c  and  then  projected  on  to  the  profile. 

32.  Modified  Equivalent  Pressure  Area  in  Inclined  Back  Dam. 
When  the  back  of  a  dam  is  inclined,  the  area  of  the  triangle  of  water 
pressure  ABC,  in  Fig.  17,  will  not  equal  the  product  of  //,  but  of  HI 
with  its  half  width,  which  latter  is  measured  parallel  to  the  base, 
consequently  the  factor   H  cannot  be  eliminated.    The  triangle 
itself  can,  however,  be  altered  in  outline  so  that  while  containing 
the  same  area,  it  will  also  have  the  vertical  height  H  as  a  factor 


Fig.    17.     Transformation  of  Inclined 

Pressure  Area  to  Equivalent 

with  Horizontal  Base 


38 


DAMS  AND  WEIRS 


^ 


DAMS  AND  WEIRS  39 

in  its  area.  This  is  effected  by  the  device  illustrated  in  Fig.  17, 
and  subsequently  repeated  in  other  diagrams.  In  this  figure  ABC 
is  the  triangle  of  water  pressure.  By  drawing  a  line  CD  parallel 
to  the  back  of  the  wall  AB,  a  point  D  is  obtained  on  the  continu- 
ation of  the  horizontal  base  line  of  the  dam.  A  and  D  are  then 
joined.  The  triangle  ABD  thus  formed  is  equal  to  ABC,  being 
on  the  same  base  AB  and  between  the  same  parallels.  The  area 

of  ABD  is  equal  to  — -X//,  and  that  of  the  wall  to  half  width 

EF  X  //.     Consequently  we  see  that   the  half   width  — -    of   the 

triangle  ABD  can  properly  represent  the  area  of  the  water  pressure, 
and  the  half  width  EF  that  of  the  wall.  The  vertical  height  H 
may,  therefore,  be  eliminated.  What  applies  to  the  whole  triangle 
would  also  apply  to  any  trapezoidal  parts  of  it.  The  direction  of  the 
resultant  line  of  water  pressure  will  still  be  as  before,  normal  to  the 
surface  of  the  wall,  i.e.,  parallel  to  the  base  BC,  and  its  incidence  on 
the  back  wTill  be  at  the  intersection  of  a  line  drawn  through  the  e.g. 
of  the  area  in  question,  parallel  to  the  base.  This  point  will  natur- 
ally be  the  same  with  regard  to  the  inclined  or  to  the  horizontally 
based  area. 

33.  Curved  Back  Profiles.  In  order  to  illustrate  the  graphical 
procedure  of  drawing  the  line  of  pressure  on  a  profile  having 
a  curved  back,  Figs.  18  and  19  are  put  forward  as  illustrations 
merely — not  as  models  of  correct  design.  In  these  profiles  the 
lower  two  laminas  of  water  pressure,  4'  and  5',  have  inclined 
bases.  Both  are  converted  to  equivalent  areas  with  horizontal 
bases  by  the  device  explained  in  the  last  section.  Take  the  lowest 
lamina  acdb;  in  order  to  convert  it  into  an  equivalent  trapezoid 
with  a  horizontal  base,  de  is  drawn  parallel  to  ac;  the  point  e  is 
joined  with  A,  the  apex  of  the  completed  triangle,  of  which  the  trap- 
ezoid is  a  portion.  When  af  is  drawn  horizontally,  the  area  acef  will 
then  be  the  required  converted  figure,  the  horizontally  measured 

half  width  of  which  multiplied  by  —  will  equal  the  area  of  the  original 

o 

trapezoid  acdb;  —  can  then  be  eliminated  as  a  common  factor  and 
5 

the  weights  of  all  the  laminas  represented  in  the  load  line  in 


40 


DAMS  AND  WEIRS 


DAMS  AND  WEIRS  41 

Fig.  18a,  by  the  half  widths  of  the  several  areas.  The  lamina  4'  is 
treated  in  a  similar  manner. 

The  graphical  processes  in  Figs.  18  and  18a  are  identical  with 
those  in  Fig.  15.  In  the  force  polygon  18a  the  water-pressure  forces 
1',  2',  3',  etc.,  are  drawn  in  directions  normal  to  the  adjoining  portion 
of  the  back  of  the  profile  on  which  they  abut,  and  are  made  equal 
in  length  to  the  half  widths  of  the  laminas  in  question.  The  back 
of  the  wall  is  vertical  down  to  the  base  of  lamina  3,  consequently 
the  forces,  1',  2',  and  3',  will  be  set  out  on  the  water-pressure  load 
line  in  Fig.  18a  from  the  starting  point,  horizontally  in  one  line. 
In  laminas  4  and  5,  however,  the  back  has  two  inclinations;  these 
forces  are  set  out  from  the  termination  of  3'  at  their  proper  direc- 
tions, i.e.,  parallel  to  their  inclined  bases  to  points  marked  a  and  b. 
The  direction  of  the  resultants  of  the  combinations,  1'  and  2',  and 
1' ',  2',  3',  will  clearly  be  horizontal.  If  Aa  and  Ab  be  joined,  then 
the  directions  of  the  combination  1'  2'  3'  4'  will  be  parallel  to  the 
resultant  line  Aa  and  that  of  1'  2'  3'  4'  5'  will  be  parallel  to  Ab. 
Thus  the  inclination  of  the  resultant  of  any  combination  of  inclined 
forces  placed  on  end,  as  in  the  water-pressure  load  line,  wrill  always 
be  parallel  with  a  line  connecting  the  terminal  of  the  last  of  the 
forces  in  the  combination  with  the  origin  of  the  load  line. 

34.  Treatment  for  Broken  Line  Profiles.  The  method  of 
ascertaining  the  relative  position  and  directions  of  the  resultants 
of  water  pressure  areas  when  the  back  of  the  wall  has  several  inclina- 
tions to  the  vertical  is  explained  as  follows:  This  system  involves 
the  construction  of  two  additional  figures,  viz,  a  force  and  ray 
polygon  built  on  the  water-pressure  load  line  and  its  reciprocal 
funicular  polygon  on  one  side  of  the  profile.  These  are  shown  con- 
structed, the  first  on  Fig.  18a,,  the  nucleus  0  of  the  vertical  force 
and  ray  polygon  being  utilized  by  drawing  rays  to  the  terminations 
of  1',  2',  3',  4',  and  5'.  In  order  to  construct  the  reciprocal  funic- 
ular polygon,  Fig.  18c,  the  first  step  is  to  find  the  c.g.'s  of  all  the 
trapezoidal  laminas  which  make  up  the  water-pressure  area,  viz, 
1'  to  5'.  This  being  done,  lines  are  drawn  parallel  to  the  bases  of 
the  laminas  (in  this  case  horizontal  lines),  to  intersect  the  back  of 
the  wall.  From  the  points  thus  obtained  the  force  lines  1',  2',  3', 
etc.,  are  drawn  at  right  angles  to  the  portions  of  the  back  of  the  wall 
on  which  they  abut.  On  these  force  lines,  which  are  not  all  parallel, 


42 


DAMS  AND  WEIRS 


the  chord  polygon  (18c)  is  constructed  as  follows:  First  the  initial 
line  AO  is  drawn  anywhere  parallel  to  its  reciprocal  AO,  in  Fig.  18a. 
From  the  intersections  of  this  line  with  the  force  line  1'  the  chord 
marked  01'  is  drawn  parallel  to  01'  in  Fig.  18a  and  intersecting 
force  line  2'.  Again  from  this  point  the  chord  01'  is  drawn  inter- 
secting force  3'  whence  the  chord  03'  is  continued  to  force  4',  and 
04'  up  to  the  force  line  5',  each  parallel  to  its  reciprocal  in  Fig.  18a. 
The  closing  line  is  05.  The  intersection  of  the  initial  and  the  closing 
lines  of  the  funicular  polygon  gives  the  position  of  the  final  resultant 
line  I7  2'  3'  4'  5',  which  is  then  drawn  from  this  point  parallel  to  its 
reciprocal  Ob  in  Fig.  18a  to  its  position  on  the  profile.  The  other 

resultants  are  obtained  in  a  sim- 
ilar manner  by  projecting  the  sev- 
eral chords  backward  till  they 
intersect  the  initial  line  OA,  these 
intersections  being  the  starting 
points  of  the  other  resultants, 
viz,  l'-4',  l'-3',  l'-2',  and  1'. 
These  resultant  lines  are  drawn 
parallel  to  their  reciprocals  in 
18a,  viz,  l'-4'  is  parallel  to  Aa, 
while  the  remainder  are  hori- 
zontal in  direction,  the  same  as 
their  reciprocals. 

This  procedure  is  identical 
with  that  pursued  in  forming  the  funicular  18b,  only  in  this  case  the 
forces  are  not  all  parallel. 

35.  Example  of  Haessler's  Method.  In  Fig.  19  the  profile 
used  is  similar  to  Fig.  18,  except  in  the  value  of  p,  which  is  2J, 
not  2.4  as  previously.  The  graphical  system  employed  is  Haessler's, 
each  lamina  as  already  described  with  reference  to  Fig.  16  being 
independently  dealt  with,  the  combination  with  the  others  taking 
place  on  the  profile  itself.  In  this  case  the  changes  of  batter  coin- 
cide with  the  divisions  of  the  laminas,  consequently  the  directions 
of  the  inclined  forces  are  normal  to  the  position  of  the  back  on  which 
their  areas  abut.  This  involves  finding  the  c.  g.  's  of  each  of  the  water- 
pressure  trapezoids,  which  is  not  necessary  in  the  first  system,  unless 
the  funicular  polygon  of  inclined  forces  has  to  be  formed.  In  spite 


Fig.  20.     Diagram  Showing  Third  Method  of 
Determining  Water  Pressure  Areas 


DAMS  AND  WEIRS  43 

of  this,  in  most  cases  Haessler's  method  will  be  found  the  handiest 
to  employ,  particularly  in  tentative  work. 

36.  Example  of  Analytical  Treatment.     In  addition  to  the 
two  systems  already  described,  there  is  yet  another  corresponding 
to  the  analytical,  an  illustration  of  which  is  given  in  Fig.  20.     In 
this  the  vertical  and  horizontal  components  of  R,  the  resultants 
(R.F.),  viz,  N  and  P  are  found.     In  this  method  the  vertical  com- 
ponent of  the  inclined  water  pressure  PI  is  added  to  the  vertical 
weight  of  the  dam  itself,  and  when  areas  are  used  to  represent 
weights  the  area  of  this  water  overlying  the  back  slope  will  have  to 
be  reduced  to  a  masonry  base  by  division  by  the  specific  gravity  of 
the  masonry. 

37.  Relations  of  R.  N.  and  W.    The  diagram  in  Fig.  20  is  a 
further  illustration  showing  the  relative  positions  of  R,  P,  PI,  N, 
and  W.    The  line  R  starts  from  a,  the  intersection  of  the  horizontal 
force  P  with  N,  the  resultant  of  all  the  vertical  forces,  for  the  reason 
that  it  is  the  resultant  of  the  combination  of  these  two  forces;  but 
R  is  also  the  resultant  of  PI  and  W,  consequently  it  will  pass  through 
a',  the  intersection  of  these  latter  forces.     The  points  a  and  a\  are 
consequently  in  the  resultant  R  and  it  follows  as  well  that  if  the 
position  of  R  is  known,  that  of  N  and  W  can  be  obtained  graphically 
by  the  intersection  of  P  or  PI  with  R.     These  lines  have  already 
been  discussed. 

UNUSUALLY  HIGH  DAMS 

38.  "High"  Dams.    An  example  will  now  be  given,  Fig.  21,  of 
the  design  of  a  high  dam,  i.e.,  one  whose  height  exceeds  the  limit 
before  stated.     As  usual  the  elementary  triangular  profile  forms 
the  guide  in  the  design  of  the  upper  portion.     We  have  seen  in 
section  24  that  the  limiting  depth  with  p  =  2.4  and  c=16  tons  = 
195  feet,  whence  for  18  tons'  limit  the  depth  will  be  219  feet.     In 
Fig.  21  the  tentative  profile  is  taken  down  to  a  depth  of  180  feet. 
The  crest  is  made  15  feet  wide  and  the  back  is  battered  1  in  30;  the 
base  width  is  made  180 X. 645  =  116  feet.    The  heel  projects  6  feet 
outside  the  axis  line.     The  graphical  procedure  requires  no  special 
explanation.     It  follows  the  analytical  in  dealing  with  the  water 
pressure  as  a  horizontal  force,  the  weight  of  the  water  overlying 
the  back  being  added  to  that  of  the  solid  dam.    For  purposes  of 


44 


DAMS  AND  WEIRS 


£uOJ.£-9f=i 


DAMS  AND  WEIRS 


45 


calculation  the  load  is  divided  into  three  parts  (1)  the  water  on 
the  sloping  back,  the  area  of  which  is  540  sq.  ft.  This  has  to  be 
reduced  by  dividing  it  by  p  and  so  becomes  225  sq.  ft.  As  tons, 
not  areas,  will  be  used,  this  procedure  is  not  necessary,  but  is  adopted 
for  the  sake  of  uniformity  in  treatment  to  avoid  errors.  The  e.g. 
of  (1)  is  clearly  2  feet  distant  from  the  heel  of  the  base  of  (3),  about 
which  point  moments  will  be  taken.  That  of  the  crest  (2)  is  13.3  feet 
and  that  of  the  main  body  (3)  obtained  by  using  formula  (7)  comes 
to  41.2  feet.  The  statement  of  moments  is  then  as  follows: 


No. 

AREA 

TONS 

LEVER  ARM 

MOMENT 

1 

225 

17' 

2 

34 

2 

360 

27 

13.3 

359 

3 

10280 

771 

41.2 

31765 

Total 

10866 

sir. 

32158 

32158 

Then  the  distance  of  N  from  the  heel  will  be- —  =  39.4  ft.;  it  thus 

815 

falls  39.4  -38.7  =  .7  ft.  within  the  middle  third.     The  distance  (/) 


between  N  and  R  is 


PH 

3N' 
II 


Now  P  =  the  area  of  the  right  angle 


triangle  whose  base  is  —  ,  or  75  feet,  and  is  6750  sq.  ft.  equivalent 
P 

6750X3     _ft_  _,,  .  ,  ,    506X180 

to  -  —  —  =  506  tons.     The  expression  then  becomes  j  =  „ 

40  oXolo 

=  37.2  feet.     The  incidence  of  R  will  then  be  37.2+39.4  =  76.6  feet 

2 
distant  from  the  heel  of  the  base.     The  —  point  being  77.3  from  the 

o 

heel,  R  falls  .7  ft.  within.  Thus  far  the  tentative  profile  has  proved 
fairly  satisfactory,  although  a  slight  reduction  in  the  base  width  is 
possible.  The  position  of  IF,  or  the  resultant  weight  of  the  portions 
2  and  3  of  the  dam  is  obtained  from  the  moment  table  already 
given,  and  is  the  sum  of  the  moments  of  (2)  and  (3)  divided  by  (2) 


+  (3)  or  =  40.2  ft.     This  falls  40.2-38.7  =  1.5  ft.  within  the 

798 


46 


DAMS  AND  WEIRS 


middle  third.    The  value  of  q  (R.F.)  is— -.7=  19.33 -.7  =  18.6 

S  /•>  _  X  •<  -I  -i    /-» 

feet,  and  m 


Then  by  formula  (Id),  N  being  815  and  P,  506 


1.96X815+V(1.96X815)2+4  (506)' 
232 


_1597+V2551367+1024144 

232 

=  15.03  tons  per  square  foot 

Extension  of  Profile.  This  value  being  well  below  the  limit 
of  18  tons  and  both  resultants  (R.F.)  and  (R.E.)  standing  within  the 
middle  third  it  is  deemed  that  the  same  profile  can  be  carried  down 
another  30  ft.  in  depth  without  widening.  The  base  length  will 
now  be  a  trifle  over  135  feet.  The  area  of  the  new  portion  (4)  is 
3769  sq.  ft.  =283  tons.  The  distance  of  its  e.g.  from  the  heel  by 
formula  (7)  is  found  to  be  63.4  feet.  The  position  of  W  will  be 
obtained  as  follows,  the  center  of  moments  being  one  foot  farther 
to  right  than  in  last  paragraph. 


No. 

TONS 

LEVER  ARM 

MOMENT 

2 

27 

14.3 

386 

3 

771 

42.2 

32536 

4 

283 

63.4 

17942 

Total 

1081 

50864 

r  rvo/?/j 

x=  =47.0  feet  from  heel  of  new  base  to  W 

1081 

As  --  is  -  -  =45  ft.,  the  incidence  of  Wi  is  2.0  ft.  within  the  base, 

o  o 

which  is  satisfactory. 

To  find  NI,  the  moment  of  the  water  on  whole  back  can  be 
added  to  that  of  W  first  obtained.     The  offset  from  the  axis  being 

210X7 
now  7  ft.,  the  area  will  be  — - —  =  735  equal  to  an  area  of  masonry 

of  —  =  306  sq.  ft.  equivalent  to  23  tons,  nearly.    The  lever  arm 

2.4: 


DAMS  AND  WEIRS  47 

being  — ,  or  2.3  feet  the  moment  about  the  heel  will  be  23X2.3  =  52.9, 

say  53.0  ft. -tons.  This  amount  added  to  the  moment  of  W\  will 
represent  that  of  Ni  and  will  be  50,864+53  =  50,917.  The  value  of 
Ni  is  that  of  Wi+  the  water  on  back  or,  1081+23  =  1104  tons. 

£OO1  *7 

The  distance  of   ATi  from  the  heel  is  then  ——=46.2  feet.     To 

1104 

689x210 

obtain  that  of  RI  the  value  of  /=- — ——  =  43.7  feet;  this  added 

o  X  L 1 U4 

i  on:  v/o 

to  46.2  =  89.9  ft.,  the  incidence  of  ft  is  therefore  -  ^^-89.9  = 

o 

90-89.9  =  .!  ft.,  within  the  middle  third  boundary.     Then  q  =  — 

6 

-.1  =22.5-. 1  =22.4  ft.,  and  m  =  1  +^  =  1  +^^  =  2.00,  nearly. 

b  loo 

To  find  c,  formula  (10)  will  be  used,  the  quantities  being  less  than 

.     „  mN    2.00X1104 

in  formula  (10i).      Here  s  = —r—  = — =16.4  tons  and  ss 

P     689 

=  —  =  ——=5.1  tons.     Then 
b      135 


l6'4  '     '^^-+(5.1)2  =  8.2+9.7  =  17.9  tons 
4 

The  limit  of  18  tons,  being  now  reached,  this  profile  will  have  to 
be  departed  from. 

39.  Pentagonal  Profile  to  Be  Widened.  The  method  now  to 
be  adopted  is  purely  tentative  and  graphic  construction  will  be 
found  a  great  aid  to  its  solution.  A  lamina  of  a  depth  of  60  ft., 
will  be  added  to  the  profile.  It  is  evident  that  its  base  width  must 
be  greater  than  that  which  would  be  formed  by  the  profile  being 
continued  down  straight  to  this  level.  The  back  batter  naturally 
will  be  greater  than  the  fore.  From  examination  of  other  profiles  it 
appears  that  the  rear  batter  varies  roughly  from  about  1  in  5  to  1 
in  8  while  the  fore  batter  is  about  1  to  1.  As  a  first  trial  an  8  ft. 
extra  offset  at  the  back  was  assumed  with  a  base  of  200  feet;  this 
would  give  the  required  front  projection.  Graphical  trial  lines 
showed  that  N  would  fall  without  the  middle  third,  and  W  as  well; 
the  stress  also  just  exceeded  18  tons  (R.E.).  A  second  trial  was 
now  made  in  which  the  back  batter  was  increased  and  the  base 
shortened  to  180  feet.  In  this  case  c  exceeded  the  18  ton  limit. 


48 


DAMS  AND  WEIRS 


Still  further  widening  was  evidently  required  at  the  heel  in  ordei 
to  increase  the  weight  of  the  overlying  water,  while  it  was  clear 
that  the  base  width  would  not  bear  reduction.  The  rear  offset 
was  then  increased  to  15  feet  and  the  base  width  to  200  feet.  The 
stresses  now  worked  out  about  right  and  the  resultants  both  fell 
within  the  middle  third.-  By  using  formula  (6)  the  distance  of  the 
e.g.  of  the  trapezoidof  water  pressure,  which  weighs  112  tons,  was 
found  to  be  7.2  feet  from  the  heel  of  the  base,  and  by  formula  (7) 
that  of  the  lowest  lamina  (5)  from  the  same  point  is  found  to  be 
91.8  feet;  the  weight  of  this  portion  is  754  tons.  These  two  new 
vertical  forces  can  now  be  combined  with  NI  whose  area  and  posi- 
tion are  known  and  thus  that  of  N2  can  be  ascertained.  NI  is  46.2 
ft.  distant  from  the  heel  of  the  upper  profile;  its  lever  arm  will, 
therefore,  be  46.2  +  15  =  61.2  feet.  The  combined  moment  about 
the  heel  will  then  be 


WEIGHT 

LEVER  ARM 

MOMENT 

Water 

112 

7.2 

806 

Ni 

1104 

61.2 

67565 

(5) 

754 

91.8 

69217 

Total 

1970 

137588 

137588 

The  incidence  of  N%  is  then  =  70  feet  from  the  heel;  as  the 

1970 

200 

middle  third  boundary  is  -— -  =  66.6  feet  distant  from  the  same  point, 

o 

N2  falls  3.4  feet  within.     The  distance  between  Nz  and  R2  (viz,  /) 
Now  P2,  or  the  horizontal  water  pressure,  has  a  reduced  area 

1139X270 
of  15187  feet,  equivalent  to  1139  tons,  consequently /=——— ^-r- 

=  52.0  feet.    This  fixes  the  incidence  of  R2  at  70+52.0  =  122.0  feet 
distant  from  the  heel;  the  §  point  is  133.3  feet  distant,  consequently 

9OO 

R2  falls  well  within  the  middle  third  and  as  q  =  122.0  — —  =  22.0  feet, 


6X22.0 


mN    1.66X1970 
—  —--  - 


tons. 


Now 


DAMS  AND  WEIRS 
P2      1139 


49 


Whence  by  formula  (10), 

16.3       |265~ 
c=~2~+\-r- +32.5  =  8.15+9.9  =  18.05  tons 


which  is  the  exact  limit  stress. 

The  value  of  s2  (the  pressure  at  the  heel)  is  obtained  by  the 

130  8 

same  formula,  using  the  minus  sign,  viz,  m  =  l — ^—^-  =  .34,  there- 

200 

mNt     .34X1970  _, 

tore,  S2=~j—  = — — — =  3.4  tons,  nearly.     Ihese  vertical  reac- 

0  ZOO 

tions  are  set  out  below  the  profile.     With  regard  to  IF2  it  is  com- 
posed of  W\-\- (5).     The  table  of  moments  is  as  follows: 


WEIGHT 

LEVER  ARM 

MOMENT 

w} 

(5) 

1081 
754 

62.0 
91.8 

67022 
69217 

W2 

1835 

136239 

The  distance  of  JF2  from  the  heel  is 


136239 
1835 


=  74.3.   JF2,  therefore, 


200 
falls   7.6  feet    within   the   middle  third;   g  =  -— -74.3  =  25.7  feet, 


and    m  =  1 
roFPj     1.77 


6X25.7 

200 
:1835 


=  1.77,    and     ra  =  l  —  .77  =  .23;    therefore,    s  = 
1835 


=  16.2  tons,  and  52  =  .23X         =  2.1  tons.    These 


b  200 

pressures  are  shown  below  the  profile. 

The  value  of  0  in  all  three  cases  is  less  than  35°  which  is  also 
one  of  the  stipulations. 

In  continuing  the  profile  below  the  270-foot  depth  the  proba- 
bility is  that  for  a  further  depth  of  50  or  60  feet  the  same  fore  and 
rear  batter  would  answer;  if  not,  the  adjustment  is  not  a  difficult 
matter  to  manipulate.  As  previously  stated,  the  incidence  of  A7 
should  be  fixed  a  little  within  the  middle  third  when  that  of  W  and 
R  will  generally  be  satisfactory. 


50  DAMS  AND  WEIRS 

In  the  force  diagram  the  water  part  of  N  is  kept  on  the  top  of 
the  load  line  W.  This  enables  the  lengths  of  the  N  series  to  be 
clearly  shown.  The  effect  is  the  same  as  if  inclined  water  pressure 
lines  were  drawn,  as  has  already  been  exhibited  in  several  cases. 

40.  Silt  against  Base  of  Dam.     In  Fig.  21,  suppose  that  the 
water  below  the  210-foot  depth  was  so  mixed  with  silt  as  to  have  a 
specific  gravity  of  1.4  instead  of  unity.     The  effect  of  this  can  be 
shown  graphically  without  alteration  of  the  existing  work.     In  the 
trapezoid  lying  between  210  and  270  the  rectangle  on  ab  represents 
the  pressure  above  210  and  the  remaining  triangle  that  of  the  lower 

60  feet  of  water.    The  base  of  the  latter,  be  is,  therefore,  —  =  7r~r  = 

p     2.4 

25  feet.     Now  the  weight  of  the  water  is  increased  in  the  proportion 

II'  X 1 .4 
of  1.4  : 1,  consequently  the  proper  base  width  will  be  —  = 

60X1  4 

—  =  35  feet.     The  triangle  acd  then  represents  the  additional 

2.  A 

pressure  area  due  to  silt.  The  normal  pressure  on  the  back  of  the 
dam  due  to  the  presence  of  silt  is  shown  graphically  by  the  triangle 
attached,  whose  base  =  cd  =  10  feet;  its  area  is  310  square  feet, 
equivalent  to  23  tons.  This  inclined  force  is  combined  with  R%  at 
the  top  right-hand  corner  of  Fig.  2 la  and  the  resultant  is  #3;  on  the 
profile  the  reciprocal  inclined  force  is  run  out  to  meet  7?2  and  from 
this  intersection  R3  can  be  drawn  up  toward  P2.  This  latter  inter- 
section gives  the  altered  position  of  Nz,  which  is  too  slight  to  be 
noticeable  on  this  scale.  The  value  of  c  and  the  inclination  of  R 
are  both  increased,  which  is  detrimental. 

If  the  mud  became  consolidated  into  a  water-tight  mass  the 
pressure  on  the  dam  would  be  relieved  to  some  extent,  as  the  earth 
will  not  exert  liquid  pressure  against  the  back.  Liquid  mud  pres- 
sure at  the  bottom  of  a  reservoir  can  consequently  be  generally 
neglected  in  design. 

41.  Filling  against  Toe  of  Dam.    Now  let  the  other  side  of  the 
dam  be  considered.     Supposing  a  mass  of  porous  material  having  an 
immersed  s.  g.  of  1 .8  is  deposited  on  the  toe,  as  is  often  actually  the 

1  8 
case.    Then  a  pressure  triangle  of  which  the  base  equals  Hy(^—  — 

45  feet  is  drawn;  its  area  will  be  1755  and  weight  132  tons;  the 


DAMS  AND  WEIRS 


51 


resultant  P\  acting  through  its  c.  g.  is  run  out  to  intersect  R%.  At  the 
same  time  from  the  lower  extremity  of  R2,  in  the  force  diagram,  a 
reciprocal  pressure  line  P±  is  drawn  in  the  same  direction  equal  in 
length  132  tons  and  its  extremity  is  joined  with  that  of  P2;  the  result- 
ing line  7?4  is  then  projected  on  the  profile  from  the  previous  inter- 
section until  it  cuts  the  force  line  P2;  this  gives  a  new  resultant  #4 
and  a  new  position  for  N,  viz,  N*,  which  is  drawn  on  the  profile ;  W  also 
will  be  similarly  affected.  The  load  on  the  toe  of  the  dam  increases 
its  stability  as  the  value  of  6  is  lessened,  the  position  of  W  is  also 
improved,  but  that  of  R±,  which  is  nearer  to  the  toe  than  R3,  is  not. 
To  adjust  matters,  the  c.  g.  of  (5)  requires  moving  to  the  right  which 

l56Tona\ 


ID  Tons- 


Ice  Pressure 


£9Z  Tony 


Fig.  22.     Diagram  Showing  Effect  of  Ice  Pressure 

is  affected  by  shifting  the  base  line  thus  increasing  the  back  and 
decreasing  the  front  batter,  retaining  the  base  length  the  same  as 
before. 

42.  Ice  Pressure.  Ice  pressure  against  the  back  of  a  dam 
has  sometimes  to  be  allowed  for  in  the  design  of  the  profile;  as  a 
rule,  however,  most  reservoirs  are  not  full  in  winter  so  that  the 
expansive  pressure  is  exerted  not  at  the  summit  but  at  some  distance 
lower  down  where  the  effect  is  negligible.  In  addition  to  this  when 
the  sides  of  a  reservoir  are  sloping,  as  is  generally  the  case,  movement 
of  ice  can  take  place  and  so  the  dam  is  relieved  from  any  pressure. 
In  the  estimates  for  the  Quaker  Bridge  dam  it  is  stated  that  an 
ice  pressure  of  over  20  tons  per  square  foot  was  provided  for.  No 


52 


DAMS  AND  WEIRS 


definite  rules  seem  to  be  available  as  to  what  allowance  is  suitable. 
Many  authorities  neglect  it  altogether. 

The  effect  of  a  pressure  of  ten  tons  per  foot  run  on  a  hundred- 
foot  dam  acting  at  the  water  level  is  illustrated  in  Fig.  22.  For 
this  purpose  a  trapezoidal  section  has  been  adopted  below  the  sum- 
mit level.  The  crest  is  made  15  feet  wide  and  10  feet  high.  This 
solid  section  is  only  just  sufficient,  as  will  appear  from  the  incidence 
of  Rf  on  the  base.  The  area  of  this  profile  is  4150  sq.  ft.,  while  one 
of  the  ordinary  pentagonal  sections  as  dotted  on  the  drawing  would 


Fig.  23.     Two  Profiles  for  Partial  Overfall  Dams 

contain  but  3325  sq.  ft.  The  increase  due  to  the  ice  pressure  is 
therefore  825  sq.  ft.  or  about  25  per  cent.  The  graphical  procedure 
hardly  needs  explanation.  The  ice  pressure  p  is  first  combined 
with  W  the  weight  of  the  dam. and  their  resultant  R  cuts  P  at  a 
point  from  which  the  final  RI  is  run  down  to  the  base  parallel  to  its 
reciprocal  in  the  force  diagram.  It  falls  just  within  the  middle 
third  of  the  base.  An  actual  example  is  given  in  section  56. 

43.  Partial  Overfall  Dams.  It  not  infrequently  happens  that 
the  crest  of  a  dam  is  lowered  for  a  certain  length,  this  portion  acting  as 
a  waste  weir,  the  crest  of  the  balance  of  the  dam  being  raised  above 
the  water  level.  In  such  cases  a  trapezoidal  outline  is  generally 
preferable  for  the  weir  portion  and  the  section  can  be  continued 
upon  the  same  lines  to  form  the  upper  part  of  the  dam,  or  the  upper 
part  can  be  a  vertical  crest  resting  on  the  trapezoidal  body.  In  a 


DAMS  AND  WEIRS  53 

trapezoidal  dam,  if  the  ratio  of  —  be  r,  the  correct  base  width  is 
obtained  by  the  following  formula: 


This  assumes  the  crest  and  summit  water  level  to  be  the  same. 
In  Fig.  23,  p  is  taken  as  2.4  and  r  as  .2.  The  base  width  with  a 

vertical  back  will  then  be  -=X  ,  =  =  50X.645X.  935  =  31.3 

Vp     V1  +  .2-.04 

feet,  and  the  crest  width  k  will  be  31  .3  X  .2  =  6.3  feet.  In  the  second 
figure  the  profile  is  shown  canted  forward,  which  is  desirable  in 
weirs,  and  any  loss  in  stability  is  generally  more  than  compensated 
for  by  the  influence  of  the  reverse  pressure  of  the  tail  water  which 
influence  increases  with  the  steepness  of  the  fore  slope  of  the  weir. 
The  base  width  is,  however,  increased  by  one  foot  in  the  second  figure. 
As  will  be  seen  in  the  next  section,  the  crest  width  of  a  weir 
should  not  be  less  than  V//+Vd;  in  this  case  77  =  45  and  d  =  5. 
This  would  provide  a  crest  width  of  6.7+2.2  =  9  feet,  which  it  nearly 
scales. 

NOTABLE  EXISTING  DAMS 

44.  Cheeseman  Lake  Dam.  Some  actual  examples  of  dam 
sections  will  now  be  exhibited  and  analyzed.  Fig.  24  is  the  section 
of  the  Cheeseman  Lake  dam  near  Denver,  Colorado,  which  is  one 
of  the  highest  in  the  world.  It  is  built  to  a  curvature  of  400  feet 
radius  across  a  narrow  canyon.  It  is  considered  a  gravity  dam, 
however,  and  will  be  analyzed  as  such.  The  section  can  be  divided 
into  three  unequal  parts  1,2,  and  3,  and  the  lines  of  pressure  (R.F.) 
and  (R.E.)  will  be  drawn  through  the  bases  of  these  three  divisions. 
Of  the  vertical  forces  (1)  has  an  area  of  756  sq.  ft.,  (2)  of  3840,  and 
(3)  of  13,356  the  total  value  of  W  being  17,952  sq.  ft.,  which  is  marked 
off  on  the  load  line  in  Fig.  24a.  With  regard  to  the  water-pressure 
areas  the  most  convenient  method,  where  half  widths  are  not  used, 
which  can  only  be  done  with  equal  divisions,  is  to  estimate  the 
areas  of  the  horizontal  pressures  only  and  set  them  off  horizontally, 
the  values  of  the  inclined  pressures  being  obtained  by  construction. 
For  this  purpose  the  triangle  of  horizontal  water  pressure  is  shown 
adjacent  to,  but  separate  from,  the  profile.  The  three  values  of  P 


54 


DAMS  AND  WEIRS 


H2 

which  are  equal  to  — —  will  be  270,  2631,  and  7636,  respectively,  the 
2p 

total  being  10,537  sq.  ft.  In  this  computation  the  value  of  p  is 
assumed  to  be  2.4.  These  several  lengths  are  now  set  out  hori- 
zontally from  the  origin  0  in  Fig.  24a,  and  verticals  drawn  upward 
intercept  the  chords,  1' ',  2',  3',  which  latter  are  drawn  from  the 
origin  0,  parallel  to  their  respective  directions,  i.e.,  normal  to  the 
adjacent  parts  of  the  wall.  The  rest  of  the  process  is  similar  to  that 
already  described,  with  reference  to  Figs.  16  and  18,  and  need  not 
be  repeated.  In  Fig.  24a  N  scales  19,450,  equal  to  1457  tons,  and 


•w 


,  |J_^7j^Jzd        (- — £=^3*93.3'— »|  o/ooo         5000  10.000  5(j.Fl. 


Fig.  24.     Profile  of  Cheeseman  Lake  Dam 


90 


on  the  profile  q  scales  15  feet,  therefore,  in  formula  (9),  m  = 

rp,       .  m  N    1.51X1457  P     783 

1.51.    1  hereiore,  s  =  — - —  = — =  12.5  tons,  and ss=  —  =  —-  = 

b  176  6      176 

4.45;  then  by  formula  (10) 


13.9  tons,  approx. 
With  regard  to  W,  q  scales  about  20  ft.,  m  then  works  out  to  1.7, 


nearly,  and  s 


mW    1.7X1346 


=  13.0  tons. 


6  176 

As  an  exercise  the  inclined  final  resultant  P  is  drawn  on  the 
profile.  This  line  is  parallel  to  Oc  in  Fig.  24a,  its  location  is  worked 
out  by  means  of  the  funicular  polygon,  the  construction  of  which 
need  not  be  explained  after  what  has  gone  before. 


DAMS  AND  WEIRS 


55 


45.  Analytical  Check.  In  order  to  check  this  result  analyt- 
ically the  procedure  will  be,  first,  calculate  the  position  of  the  e.g.  of 
the  trapezoids  (2)  and  (3)  relative  to  the  rear  corner  of  their  bases 
by  formula  (7)  and  also  the  positions  of  the  resultants  of  the  vertical 
components  of  the  water  pressure  overlying  the  back  with  regard 
to  the  same  points  by  formula  (6).  Second,  convert  the  areas  into 
tons  by  multiplying  by  ^V  The  statement  of  moments  about  the 
heel  of  the  base,  with  the  object  of  finding  the  position  of  W  is  given 
below. 


Moment  of  (1) 

56.7  X32.5 

1843 

Moment  of  (2) 

288     X47.9 

=  13795 

Moment  of  (3) 

1001     X75 

=  75075 

Total   IF  = 

1346  tons 

90713 

The  distance  of  IF  from  the  heel  will  then  be  '• —  —  =  67.5  ft.     In 

1346 

order  to  obtain  A",  the  moments  of  the  water  weights  will  have  to 
be  added  as  below. 


Moment  of  W 

1346X67.5 

=  90713 

Moment  of  w\ 

10X21.6 

=  216 

Moment  of  Wz 

107X9 

=  963 

Total  N  = 

1463  tons 

91892 

and 


To  find  the  incidence  of  R  and  its  distance  (q)  from  the  center 
point,  that  from  the  known  position  of  N  must  be  computed  from 


the  formula  /= 
176 


3N      3X1463 


=  ^  ft.,  therefore,  q=  (62.8+40.0) 


-—  =  14.8  feet.     This  is  close  to  the  value  obtained  graphically 

which  was  taken  as  15  feet.     The  value  of  N  is  also  seen  to  be  close 
to  that  obtained  graphically.     The  value  of  q  with  regard  to  W  (R.E.) 

is  as  follows,  q  =  — —  67.5  —  20.5  feet,  almost  exactly  what  it  scales  on 


56 


DAMS  AND  WEIRS 


the  diagram.     In  this  profile  the  upper  part  is  light,  necessarily  made 
up  for  in  the  lower  part. 

At  the  upper  base  line  of  (2)  the  incidence  of  W  is  exactly  at 
the  middle  third  edge,  while  R  falls  within  it.  At  the  final  base  the 
position  of  N  is  62.8  distant  from  the  heel  and  the  inner  third  point 

1  T C^. 

=  58.6  distant,  consequently  the  incidence  of  N  lies  4.2  feet 


is 


3 


within  the  boundary. 

If  the  position  of  N  were  made  obligatory  at  the  inner  edge  of 
the  middle  third,  the  value  of  W  would  be  increased,  but  R  would 


_L  RADIUS  4/0'  *-z3.£ 


(o) 


(b) 


O  IO?03Qii050  tOO  /SOFt: 

0  IOOOZOOO  5000  100003.  Ft. 


Fig.  25.     Profile  of  Roosevelt  Dam  across  Salt  River,  Arizona 

be  decreased.  There  may  have  been  special  reasons  for  limiting  the 
maximum  stress  (R.E.).  On  Fig.  24  the  position  of  N  is  obtained 
by  the  intersection  of  the  horizontal  resultant  P  with  R  prolonged 
upward.  If  the  stress  were  calculated  on  the  supposition  that  the 
structure  was  an  arched  dam,  it  would  amount  to  21 J  tons  by  the 
"long"  formula,  given  in  section  78,  Part  II. 

46.  Roosevelt  Dam.  In  Fig.  25  is  given  the  profile  of  the 
Roosevelt  dam,  and  Fig.  26  is  the  site  plan  of  that  celebrated  work. 
For  some  years,  the  Roosevelt  dam  was  the  highest  gravity  dam  in 
existence.  It  spans  a  very  deep  canyon  of  the  Salt  River  in  Arizona 


DAMS  AND  WEIRS 


57 


and  impounds  the  enormous  quantity  of  1 J  million  acre-feet  of  water, 
which  will  be  utilized  for  irrigation.  This  work  is  part  of  one  of 
the  greatest  of  the  several  large  land  reclamation  projects  under- 
taken by  the  U.  S.  Government  for  the  watering  and  settling  of 
arid  tracts  in  the  dry  zone  of  the  western  states. 

The  profile  is  remarkable  for  the  severe  simplicity  of  its  out- 
line. It  closely  follows  the  elementary  profile  right  down  to  its 
extreme  base  and  forms  a  powerful  advocate  for  this  simple  style 
of  design.  The  graphical  procedure  is  similar  to  that  in  the  last 
example.  The  section  is  divided  into  three  divisions.  As  the 
first  two  are  comparatively  small,  the  triangle  of  forces  in  Fig.  25a 


Fig.  26.     Site  Plan  for  Roosevelt  Dam 

is  first  plotted  at  a  large  scale  in  pencil  and  the  inclinations  of  the 
resultants  thus  obtained  are  transferred  to  the  profile;  this  accounts 
for  the  long  projecting  lines  near  the  origin  of  the  force  diagram 
which  also  appear  in  some  previous  examples.  A  neater  method 
for  overcoming  this  difficulty  is  that  adopted  in  the  next  figure, 
when  the  forces  (1)  and  (2)  are  first  amalgamated  into  one  before 
being  plotted  on  the  force  diagram. 

In  Fig.  25a,  N  scales  roughly  19,000  sq.  ft.,  equivalent  to  1425 

120 

tons,  q  also  measures  approximately  20  ft.,  then  ra  =  l+— -  =  1.75, 

loU 


and  s  = 


mN    1425X1.75 


160 


i  K  K  4.  P     826 

=  15.5  tons.     ss=  f-  =  - 


formula  (10) 


58  DAMS  AND  WEIRS 


=  17  tons  roughly 

With  regard  to  W,  q  measures  23   ft.  and  m  works  out  to  1.86 

.  mW    1.86X1378  ., 

therefore  s  =— —  =  —  -  =16  tons  per  sq.  ft. 

o  160 

This  dam  is  built  on  a  radius  of  410  feet,  measured  from  the 
axis;  if  measured  from  the  extrados  of  the  curve  at  the  base  it  will 
be  420  feet  and  the  arch  stress  as  calculated  from  the  "long"  formula 
used  in  "Arched  Dams"  will  amount  to  23.3  tons. 

The  site  plan  given  in  Fig.  26  forms  an  instructive  example  of  the 
arrangement  of  spillways  cut  in  the  solid  rock  out  of  the  shoulders 
of  the  side  of  the  canyon,  the  material  thus  obtained  being  used  in 
the  dam.  These  spillways  are  each  200  feet  wide  and  are  excavated 
down  to  five  feet  below  the  crest  of  the  dwarf  waste  weir  walls  which 
cross  them.  This  allows  of  a  much  greater  discharge  passing  under 
a  given  head  than  would  be  the  case  with  a  simple  channel  without 
a  drop  wall  and  with  bed  at  the  weir  crest  level.  The  heading  up, 
or  afflux,  is  by  this  means  diminished  and  that  is  a  matter  affecting 
the  height  given  to  the  dam  crest. 

47.  New  Croton  Dam.  The  profile  of  the  New  Croton  dam 
constructed  in  connection  with  the  water  supply  of  New  York  City 
is  given  in  Fig.  27.  This  dam  has  a  straight  alignment  and  is  1168 
feet  long.  Waste  flood  water  is  accommodated  by  an  overfull  weir 
1000  ft.  in  length,  which  is  situated  on  one  flank  forming  a  continua- 
tion of  the  dam  at  right  angles  to  its  axis.  The  surplus  water 
falls  into  the  Rocky  River  bed  and  is  conveyed  away  by  a  separate 
channel.  An  elevation  and  plan  of  this  work  are  given  in  Figs. 
28  and  29. 

The  system  of  graphical  analysis  employed  in  this  case  is  differ- 
ent from  that  in  the  last  two  examples  and  is  that  illustrated  in  Fig. 
18,  where  independent  combinations  of  vertical  and  inclined  forces 
are  used.  The  profile  is  divided  into  four  divisions,  the  first  being 
a  combination  of  two  small  upper  ones.  The  further  procedure  after 
the  long  explanations  already  given  does  not  require  any  special 
notice  except  to  point  out  that  the  directions  of  the  combined  forces 
1',  l'+2',  l'+2'+3'  etc.,  in  (d)  are  drawn  parallel  to  their  reciprocal 
lines  on  Fig.  27a,  namely  to  the  chords  Oa,  Ob,  Oc,  and  Od,  respec- 
tively. The  final  resultants  are  A  (R.F.)  and  W  (R.E.).  The 


DAMS  AND  WEIRS 


59 


value  of  W  is  1380  tons  and  that  of  N  is  1484  tons,  consequently 
applying  formula  (10),  q  in  the  first  case  scales  26  feet  and  m  works 


out  to  1.82  therefore  s  = 


iyu 


W,  s  and  c  are  identical. 
regard  to 

1'22Xl484 


=  13.2  tons  =c,  as  with 


With  regard  to  N,  q  scales  7  feet,  consequently  ra  =  1  +7^7; 


L22,    g  = 


J.  c/vJ 


tons    only.     As    P  =  10,010  =  750    tons, 


P  95 

—  =  4  tons;  therefore  c  =  -^- 


(9  5) 


2  =  ll  tons,  which  is  very 


moderate.     It  is  probable  that  other  external  pressures  exist  due 

4,£ 

1700       &£3       H-Z?0 


(c) 

Fig.  27.     Diagram  of  Profile  of  New  Croton  Dam  Showing  Influence  Lines  as  in  Fig.  18 

to  filling  in  front  and  rear,  as  also  ice  pressure,  which  would  materi- 
ally modify  the  result  above  shown.  This  dam,  like  the  Cheese- 
man,  is  of  the  bottleneck  profile,  it  is  straight  and  not  curved  on  plan. 
48.  Assuan  Dam.  The  section,  Fig.  30,  is  of  the  Assuan  dam 
in  Egypt,  which  notable  work  was  built  across,  the  Nile  River  above 
the  first  cataract.  As  it  stands  at  present  it  is  not  remarkable  for 


DAMS  AND  WEIRS 


61 


its  height,  but  what  it  lacks  in  that  respect,  as  in  most  eastern 
works,  is  made  up  in  length,  which  latter  is  6400  feet.  No  single 
irrigation  work  of  modern  times  has  been  more  useful  or  far-reaching 
in  beneficial  results  upon  the  industrial  welfare  of  the  people  than 
this  dam.  Its  original  capacity  was  863,000  acre-feet  and  the  back 
water  extended  for  140  miles  up  the  river.  The  work  is  principally 
remarkable  as  being  the  only  solid  dam  which  passes  the  whole 
discharge  of  a  large  river  like  the  Nile,  estimated  at  500,000  second- 
feet,  through  its  body,  for  which  purpose  it  is  provided  with  140 
low  and  40  high  sluices.  These  are  arranged  in  groups  of  ten,  each 


-Jff' 


PRESENT  EL. 


STEEL    RODS  CO/ittECTING 
HEW  WORK  WITH  OLD 


Fig.  30.     Assuan  Dam  across  the  Nile  Showing  Old  and  New  Profiles 

low  sluice  is  23  feet  deep  by  6J  feet  wide  with  the  dividing  piers 
16 J  feet  wide.  The  diminution  of  the  weight  of  the  dam  due  to 
sluices  necessitates  an  excess  of  width  over  what  would  be  sufficient 
for  a  solid  dam;  in  addition  to  which  the  maximum  pressure  in  the 
piers  is  limited  to  the  extremely  low  figure  of  5  tons  of  2000  pounds. 
The  designers  have  thus  certainly  not  erred  on  the  side  of  boldness; 
the  foundation  being  solid  granite,  would  presumably  stand,  with 
perfect  safety,  pressure  of  treble  that  intensity,  while  the  masonry, 
being  also  granite,  set  in  cement  mortar,  is  certainly  capable  of 
carrying  a  safe  pressure  of  15  tons,  as  many  examples  prove. 


EASTEfftt   DESERT   HILLS 


DAMS  AND  WEIRS 


63 


This  dam  has  proved  such  a  financial  success  that  it  has  recently 
been  raised  by  23  feet  to  the  height  originally  projected.  The 
water  thus  impounded  is  nearly  doubled  in  quantity,  i.e.,  to 
over  1|  million  acre-feet;  exceeding  even  that  of  the  Salt  River 
reservoir^in  Arizona.  As  it  was  decided  not  to  exceed  the  low  unit 
pressure  previously  adopted,  the  profile  has  been  widened  by  16J 
feet  throughout.  A  space  has  been  left  between  the  new  and  the 
old  work  which  has  been  subsequently  filled  in  with  cement  grout 
under  pressure,  in  addition  to  which  a  series  of  steel  rods  has  been  let 


Fig.  32.     View  of  Assuan  Dam  before  Being  Heightened  with  Sluices  in  Operation 

into  the  old  face  by  boring,  and  built  into  the  new  work.  The  enlarge- 
ment is  shown  in  the  figure.  The  sluices  are  capable  of  discharging 
500,000  second  feet;  as  their  combined  area  is  25,000  square  feet 
this  will  mean  a  velocity  of  20  feet  per  second.  Owing,  however,  to 
the  possibility  of  adjustment  of  level,  by  manipulation  of  the  sluice 
gates,  they  will  never  be  put  to  so  severe  a  test. 

A  location  plan  and  longitudinal  section  shown  in  Fig.  31,  a 
view  of  the  sluices  in  operation,  Fig.  32,  and  a  view  of  the  new  work 
in  process,  Fig.  33,  will  give  a  good  idea  of  the  construction  features. 


64 


DAMS  AND  WEIRS 


DAMS  AND  WEIRS 


65 


49.  Cross  River  and  Ashokan  Dams.  Two  further  sections  are 
given  in  Figs.  34  and  35,  the  first  of  the  Cross  River  dam,  and  the 
second  of  the  Ashokan  dam  in  New  York.  Both  are  of  unusually 
thick  dimensions  near  the  crest,  this  being  specially  provided  to 
enable  the  dams  to  resist  the  impact  of  floating  ice.  These  profiles 
are  left  to  be  analyzed  by  the  student.  The  Ashokan  dam  is  pro- 
vided with  a  vertical  line  of  porous  blocks  connected  with  two  inspec- 
tion galleries.  This  is  a  German  innovation,  which  enables  any 


Fig.  34.     Profile  of  Cross  River  Dam 


Fig.  35.     Profile  of  Ashokan  Dam 


leakage  through  the  wall  to  be  drained  off,  thereby  guarding  against 
hydrostatic  uplift.    This  refinement  is  now  frequently  adopted. 

50.  Burrin  Juick  Dam.  The  Burrin  Juick  dam  in  Australia, 
Fig.  36,  which  is  generally  termed  "Barren  Jack",  is  a  close  copy  of 
the  Roosevelt  dam,  Fig.  25,  and  is  a  further  corroboration  of  the 
excellence  of  that  profile.  It  is  built  across  the  Murrumbidgee 
River  in  New  South  Wales  not  far  from  the  new  Federal  Capital. 
Its  length  is  784  feet  on  the  crest,  the  maximum  height  being  240 
feet.  The  fore  batter  is  3  vertical  to  2  horizontal,  and  the  back 
batter  20  vertical  to  1  horizontal,  both  identical  with  those  adopted 
in  the  Roosevelt  dam;  the  crest  width  is  18  feet.  It  is  built  on  a 
curve  to  a  radius  of  1200  feet.  This  dam  will  impound  785,000 
acre-feet.  The  material  of  which  the  dam  is  composed  is  crushed 
sandstone  in  cement  mortar  with  a  plentiful  sprinkling  of  large 
"plums"  of  granite.  The  ultimate  resistance  of  specimen  cubes 


66 


DAMS  AND  WEIRS 


was  180  "long"  tons,  per  square  foot;  the  high  factor  of  safety  of 
12  was  adopted,  the  usual  being  8  to  10.  The  maximum  allowable 
stress  will,  therefore,  reduce  to  15  "long"  tons  =  16.8  American 
short  tons. 

With  regard  to  the  maximum  stresses,  for  Reservoir  Full,  N  = 
16,100,  equivalent  to  1210  tons,  and  q  scales  about  15  feet,  conse- 


15580 


I  u ^£%,5- 4 jj  & 

•Z&X&Ztl^^^^/^y/X^^j.  '^/^ZZ^^/s'sztt'//^     f 


Fig.  36.     Analytical  Diagram  Showing  Profile  of  Burrin  Juick  Dam  in  Australia 

quently  m  comes  to  1.62,  and  s  =  ~ r~ =~  "-.A^  —=13.5  tons,  and 

o  14o 

—=——=4.8  tons.     Whence 
o      145 


=  15  tons 


For  Reservoir  Empty,  W=  15,580  feet  or  1170  tons,  ?  =  24,  m  =  2 
^  =  ^7-=     145     =16  tons' near1^ 


DAMS  AND  WEIRS 


67 


The  above  proves  that  the  stress  (R.E.)  is  greater  than  that  of 
(R.F.).  Probably  allowance  was  made  for  masses  of  porous  filling 
lying  at  the  rear  of  the  dam,  which  would  cause  N  and  W  to  be 
shifted  forward  and  so  equalize  the  pressure.  It  will  be  noticed 
that  the  incidence  of  N,  tne  vertical  component  (R.E.)  falls  exactly 
at  the  edge  of  the  middle  third,  a  condition  evidently  observed  in 
the  design  of  the  base  width. 


5q.f!.=58Tons 


Fig.  37.     Profile  of  Arrow  Rock  Dam,  Idaho,  Showing  Incidence  of  Centers  of  Pressure  on  Base 

The  dam  is  provided  with  two  by- washes  400  feet  wide;  the 
reservoir  will  be  tapped  by  a  tunnel  14X13  feet,  the  entrance 
sluices  of  which  will  be  worked  from  a  valve  tower  upstream,  a 
similar  arrangement  to  that  in  the  Roosevelt  dam.  It  is  interesting 
to  note  that  an  American  engineer  has  been  put  in  charge  of  the 
construction  of  this  immense  work  by  the  Commonwealth  Govern- 
ment. 

51.  Arrow  Rock  Dam.  The  highest  dam  in  the  world  now 
just  completed  (1915)  is  the  Arrow  Rock  on  the  Boise,  Idaho, 
project,  a  U.  S.  reclamation  work.  From  the  crest  to  the  base  the 
fore  curtain  is  351  feet.  A  graphical  analysis  of  the  stress  in  the 


68 


DAMS  AND  WEIRS 


base  is  given  in  Fig.  37.     For  Reservoir  Empty,  JF  =  2609  tons,  and 

228  2W 

q  measures  38  feet;  therefore  m  =  1  +  -- —  =  2  nearly  and  s  =  —  = 

222  b 


2X-^  ^  =  23.5  tons.  .  For  Reservoir  Full,  q  =  27,  and  m  =  l+—  = 

222  222  • 

1.73X2609     9n  P     1610     7  _  20.2  , 

— 222 =20.2  tons.     *a=y  =  -^-  =  7.3  '  C  =  ~2~  + 


(20. 2)2 

— h(7.3)2  =  22.6  tons.     These   values   are,   of   course,   but 

approximate. 

Thus  the  compressive  stresses  (R.F.)  and  (R.E.)  are  practically 
equal,  and  the  incidence  of  W  and  also  of  N  is  close  to  the  edge  of 


Seclionflfl 


Fig.  38.     Location  Plan  of  Arrow  Rock  Dam 
Courtesy  of  "Engineering  Record11 

the  middle  third.  The  dam  is  built  on  a  radius  of  661  feet  at  the 
crest.  The  high  stresses  allowed  are  remarkable,  as  the  design  is  on 
the  gravity  principle,  arch  action  being  ignored.  The  curvature 
doubtless  adds  considerably  to  safety  and  undoubtedly  tends  to 
reduce  the  compressive  stresses  by  an  indeterminate  but  substantial 
amount.  It  is  evident  that  formula  (10)  has  been  applied  to  the 


DAMS  AND  WEIRS 


G9 


design.  Reference  to  Figs.  38  and  39  will  show  that  the  dam  is 
divided  into  several  vertical  sections  by  contraction  joints.  It  is 
also  provided  with  inspection  galleries  in  the  interior  and  vertical 
weeper  drains  10  feet  apart.  These  intercept  any  possible  seepage, 
which  is  carried  to  a  sump  and  pumped  out.  These  precautions  are 


Earth  Surface 


(jra* 


Diversion  Tunnel 


Fig.  39.     Elevation  of  Arrow  Rock  Dam 

to  guard  against  hydrostatic  uplift.     The  simplicity  of  the  outline, 
resembling  that  of  the  Roosevelt  dam,  is  remarkable. 

SPECIAL  FOUNDATIONS 

52.  Dams  Not  Always  on  Rock.  Dams  are  not  always  founded 
on  impervious  rock  but  sometimes,  when  of  low  height,  are  founded 
on  boulders,  gravel,  or  sand.  These  materials  when  restrained 
from  spreading,  and  with  proper  arrangements  to  take  care  of  sub- 
percolation,  are  superior  to  clay,  which  latter  is  always  a  treacher- 
ous material  to  deal  with.  When  water  penetrates  underneath  the 
base  of  a  dam,  it  causes  hydrostatic  uplift,  which  materially  reduces 
the  effective  weight  of  the  structure.  Fig.  40  represents  a  wall 
resting  on  a  pervious  stratum  and  upholding  water.  The  water 
has  ingress  into  the  substratum  and  the  upward  pressure  it  will 
exert  at  c  against  the  base  of  the  wall  will  be  that  due  to  its  depth, 
in  this  case  30  feet.  Now  the  point  of  egress  of  the  percolation  will 
be  at  6,  and,  as  in  the  case  of  a  pipe  discharging  in  the  open,  pressure 
is  nil  at  that  point;  consequently  the  uplift  area  below  the  base  will 

Hxb 


be  a  triangle  whose  area  equals 


The  diagram,  Fig.  40,  shows 


the  combinations  of  the  horizontal  water  pressure  P  with  the  hydro- 
static uplift  V  and  with  the  weight  of  the  wall  W.  P  is  first  com- 
bined with  V,  RI  resulting,  whose  direction  is  upward.  RI  is  then 


70 


DAMS  AND  WEIRS 


combined  with  W,  R2  being  their  resultant.  The  conditions  with- 
out uplift  are  also  shown  by  the  dotted  line  drawn  parallel  to  dc  in 
Fig.  40.  The  line  ab  is  termed  the  hydraulic  gradient;  it  is  also  the 
piezometric  line,  i.e.,  a  line  connecting  water  levels  in  piezometer 
tubes,  were  such  inserted. 

Fig.  41  shows  the  same  result  produced  on  the  assumption  that 
the  portion  of  the  wall  situated  below  the  piezometric  line  is  reduced 
in  weight  by  an  equal  volume  of  water,  i.e.,  the  s.g.  of  this  part  may 
be  assumed  reduced  by  unity,  i.e.,  from  2.4  to  1.4.  The  wall  is 


Fig.  40.     Effect  of  Uplift  on  Dam  Shown  Graphically 

thus  divided  diagonally  into  two  parts,  one  of  s.g.  2.4  and  the  other 
of  s.g.  1.4.  The  combination  of  l-f-2  with  P  is  identical  in  result 
with  that  shown  in  Fig.  40.  In  the  subsequent  section,  dealing 
with  "Submerged  Weirs  on  Sand",  this  matter  of  reduction  in  weight 
due  to  notation  is  frequently  referred  to. 

53.  Aprons  Affect  Uplift.  Fig.  42  is  further  illustrative  of  the 
principle  involved  in  dams  with  porous  foundations.  The  pentag- 
onal profile  abc,  is  of  sufficient  base  width,  provided  hydrostatic 
uplift  is  absent.  Supposing  the  foundation  to  be  porous,  the  area 


DAMS  AND  WEIRS 


71 


of  uplift  will  be  aj)c,  in  which  bai,  equals  ab.  This  area  is  equal 
to  abc,  consequently  practically  the  whole  of  the  profile  lies  below 
the  hydraulic  gradient,  may  be  considered  as  submerged,  and  hence 
loses  weight;  its  s.g.  can  thus  be  assumed  as  reduced  by  unity,  i.e., 
from  p  to  p  —  1.  The  correct  base  width  will  then  be  found  by 

TT  TT 

making  b=  instead  of  —==     The  new  profile  will  then  be  adb; 

Vp-1  Vp 

the  base  width  having  been  thus  extended,  the  uplift  is  likewise 
increased  in  the  same  proportion.  Now  supposing  an  impervious 
apron  to  be  built  in  front  of  the  toe  as  must  be  the  case  with  an 
overfall  dam;  then  the  area  of  uplift  becomes  barf,  and  the  piezo- 


P-450 


Fig.  41.     Diagram  Showing  Identical  Result  If  Weight  Is  Considered  Reduced 
Due  to  Submersion 

metric  line  and  hydraulic  gradient,  which  in  all  these  cases  happen 
to  be  one  and  the  same  line,  is  ae.  Under  these  circumstances  the 
comparatively  thin  apron  is  subjected  to  very  considerable  uplift 
and  will  blow  up  unless  sufficiently  thick  to  resist  the  hydrostatic 
pressure.  The  low  water,  or  free  outlet  level,  is  assumed  to  be  at 
the  level  e,  consequently  the  fore  apron  lies  above  this  level  and  is 
considered  as  free  from  flotation  due  to  immersion. 

54.  Rear  Aprons  Decrease  Uplift.  Another  case  will  now  be 
examined.  In  Fig.  42  suppose  the  fore  apron  removed  and  a  rear 
apron  substituted.  In  this  case  the  point  of  ingress  of  the  percolat- 
ing water  is  thrown  back  from  b  to  b'  the  hydraulic  gradient  is  a'c, 


72 


DAMS  AND  WEIRS 


the  triangle  of  hydrostatic  uplift  is  b'azC.  This  uplift  from  b'  to  b 
is  more  than  neutralized  by  the  rectangle  of  water  a'abb',  which 
overlies  the  rear  apron;  the  latter  is  therefore  not  subject  to  any 
uplift  and,  owing  to  its  location,  is  generally  free  from  erosion  by 
moving  water,  consequently  it  can  be  made  of  clay,  which  in  this 
position  is  water-tight  as  concrete  masonry.  A  glance  at  Fig.  42 
will  demonstrate  at  once  the  great  reduction  of  uplift  against  the 
base  of  the  wall  effected  by  the  expedient  of  a  rear  apron,  the  uplift 
being  reduced  from  ajbc  to  fbc,  more  than  one-half.  Thus  a  rear 


Fig.  42.     Diagram  Showing  Uplift  with  and  without  Fore  and  Rear  Aprons 

apron  is  a  sure  remedy  for  uplift  while  the  fore  apron,  if  solid,  should 
be  made  as  short  as  possible,  or  else  should  be  formed  of  open  work, 
as  heavy  slabs  with  open  joints.  In  the  rear  of  overfall  dams  stanch- 
ing clay  is  often  deposited  by  natural  process,  thus  forming  an 
effective  rear  apron.  Many  works  owe  their  security  to  this  fact 
although  it  often  passes  unrecognized. 

55.  Rock  Below  Gravel.  Fig.  43  represents  a  dam  founded 
on  a  stratum  of  pervious  material  beneath  which  is  solid  rock.  A 
fore  curtain  wall  is  shown  carried  down  to  the  impervious  rock. 
The  conditions  now  are  worse  than  those  resulting  from  the  imper- 


DAMS  AND  WEIRS 


73 


vious  fore  apron  in  Fig.  42  as  the  hydraulic  gradient  and  piezometric 
line  are  now  horizontal.  The  reduced  area  of  vertical  hydrostatic 
pressure  is  1006  against  which  the  wall  can  only  furnish  1200;  there 
is,  therefore,  an  effective  area  of  only  134  to  resist  a  water  pressure 
at  the  rear  of  800,  consequently  the  wall  must  fail  by  sliding  or 
overturning  as  the  graphical  stress  lines  clearly  prove.  The  proper 
position  of  a  diaphragm  curtain  wall  is  at  the  heel,  not  at  the  toe  of 
the  dam;  in  this  location  it  will  effectively  prevent  all  uplift.  In 
the  case  where  an  impervious  stratum  does  not  occur  at  a  reasonable 
depth  the  remedy  is  to  provide  a  long  rear  apron  which  will  reduce 
hydrostatic  uplift  to  as  small  a  value  as  may  be  desired,  or  else  a 
combination  of  a  vertical  diaphragm  with  a  horizontal  apron  can  be 


LI  HE 


SOLIO       HOCK 
Fig.  43.     Effect  of  Impervious  Fore  Curtain  Wall  on  Uplift 

used.  In  many  cases  a  portion  only  of  the  required  rear  apron  need 
be  provided  artificially.  With  proper  precautionary  measures  the 
deposit  of  the  remaining  length  of  unfinished  apron  can  safely  be 
left  for  the  river  to  perform  by  silt  deposit,  if  time  can  be  afforded 
for  the  purpose. 

56.  Gravity  Dam  Reinforced  against  Ice  Pressure.  This  sec- 
tion will  be  concluded  with  a  recent  example  of  a  gravity  dam  rein- 
forced against  ice  pressure,  which  is  given  in  Fig.  44,  viz,  that  of  the 
St.  Maurice  River  dam  situated  in  the  Province  of  Quebec.  The 
ice  pressure  is  taken  as  25  tons  per  foot  run,  acting  at  a  level  corre- 
sponding to  the  crest  of  the  spillway,  which  latter  is  shown  in  Fig. 
58.  The  profile  of  Fig.  44  is  pentagonal,  the  crest  has  been  given 


74 


DAMS  AND  WEIRS 


the  abnormal  width  of  20  feet,  while  the  base  is  f  of  the  heigt 
which  is  about  the  requirement,  were  ice  pressure  not  considere 
The  horizontal  ice  pressure,  in  addition  to  that  of  the  water  uphel 
will  cause  the  line  of  pressure  to  fall  well  outside  the  middle  thir 
thus  producing  tension  in  the  masonry  at  the  rear  of  the  sectio 
To  obviate  this,  the  back  of  the  wall  is  reinforced  with  steel  rods 
the  extent  of  1J  square  inches  per  lineal  foot  of  the  dam.  If  tl 
safe  tensile  strength  of  steel  be  taken  at  the  usual  figure  of  16,0( 


Fig.  44.     Profile  of  Saint  Maurice  River  Dam  at  Quebec 

pounds,  or  8  tons  per  square  inch,  the  pull  exerted  by  the  reinforc 
ment  against  overturning  will  be  12  tons  per  foot  run.    This  for 
can  be  considered  as  equivalent  to  a  load  of  like  amount  applied 
the  back  of  the  wall,  as  shown  in  the  figure.    The  section  of  tl 
dam  is  divided  into  two  parts  at  El  309  and  the  incidence  of  the 
resultant  pressure  at  this  level  and  at  the  base  is  graphically  obtained 
The  line  of  pressure  connecting  these  points  is  drawn  on  the  pro- 
file.   The  line  falls  outside  the  middle  third  in  the  upper  half  oi- 
the  section  and  within  at  the  base,  the  inference  being  that  the 


DAMS  AND  WEIRS  75 

section  would  be  improved  by  conversion  into  a  trapezoidal  outline 
with  a  narrower  crest  and  with  some  reinforcement  introduced  as 
has  been  done  in  the  spillway  section,  shown  in  Fig.  58. 

It  will  be  noticed  that  the  reinforcement  stops  short  at  El 
275.  This  is  allowed  for  by  assuming  the  imposed  load  of  12  tons 
removed  at  the  base  of  the  load  line  in  the  force  polygon.  The 
line  7?5  starting  from  the  intersection  of  7?4  with  a  horizontal  through 
El  275.0  is  the  final  resultant  at  the  base.  This  example  is  most 
instructive  as  illustrating  the  combination  of  reinforcement  with  a 
gravity  section  in  caring  for  ice  pressure,  thus  obviating  the  undue 
enlargement  of  the  profile. 

GRAVITY  OVERFALL  DAMS  OR  WEIRS 

57.  Characteristics  of  Overfalls.  When  water  overflows  the 
crest  of  a  dam  it  is  termed  an  overfall  dam  or  weir,  and  some  modi- 
fication in  the  design  of  the  section  generally  becomes  necessary. 
Not  only  that,  but  the  kinetic  effect  of  the  falling  water  has  to  be 
provided  for  by  the  construction  of  an  apron  or  floor  which  in  many 
cases  forms  by  far  the  most  important  part  of  the  general  design. 
This  is  so  pronounced  in  the  case  of  dwarf  diversion  weirs  over  wide 
sandy  river  beds,  that  the  weir  itself  forms  but  an  insignificant  part 
of  the  whole  section.  The  treatment  of  submerged  weirs  with  aprons, 
will  be  given  elsewhere.  At  present  the  section  of  the  weir  wall 
alone  will  be  dealt  with. 

Typical  Section.  Fig.  45  is  a  typical  section  of  a  trapezoidal 
weir  wall  with  water  passing  over  the  crest.  The  height  of  the  crest 
as  before,  will  be  designated  by  H,  that  of  reservoir  level  above 
crest  by  d,  and  that  of  river  below  by  D.  The  total  height  of  the 
upper  still  water  level,  will  therefore,  be  H-\-d. 

The  depth  of  water  passing  over  the  crest  should  be  measured 
some  distance  upstream  from  the  overfall  just  above  where  the 
break  takes  place;  the  actual  depth  over  the  crest  is  less  by  reason 
of  the  velocity  of  the  overfall  being  always  greater  than  that  of 
approach.  This  assumes  dead  water,  as  in  a  reservoir,  in  the  upper 
reach.  On  a  river  or  canal,  however,  the  water  is  in  motion  and  has 
a  velocity  of  approach,  which  increases  the  discharge.  In  order  to 

allow  for  this,  the  head  (h)  corresponding  to  this  velocity,  or  — — 


70 


DAMS  AND  WEIRS 


multiplied  by  1.5  to  allow  for  impact,  or  /i  =  .0233F2,  should  be 
added  to  the  reservoir  level.  Thus  supposing  the  mean  velocity 
of  the  river  in  flood  to  be  10  feet  per  second  100 X. 0233  or  2.3 
feet  would  have  to  be  added  to  the  actual  depth,  the  total  being 
15  feet  in  Fig.  45.  The  triangle  of  water  pressure  will  have  its 
apex  at  the  surface,  and  its  base  will,  for  the  reasons  given 
previously,  be  taken  as  the  depth  divided  by  the  specific  gravity 
of  the  material  of  the  wall.  The  triangle  of  water  pressure  will 


Fig.  45.     Typical  Section  of  Trapezoidal  Weir  Wall 

be  truncated  at  the  crest  of  the  overfall.     The  water  pressure  acting 
against  the  back  of  the  wall  will  thus  be  represented  by  a  trapezoid, 


not  a  triangle,  whose  base  width  is 


H+d 


and  its  top  width  at  crest 


level — .     Its  area  therefore  (back  vertical)  will  be  I  -       — I — '-  1— . 
P  \    P         PS* 

If  the  back  is  inclined  the  side  of  the  trapezoid  becomes  HI.     The 
general  formula  is  therefore 


DAMS  AND  WEIES  77 

HI  being  the  inclined  length  of  the  back  of  the  wall.     The  vertical 
distance  of  its  point  of  application  above  the  base  according  to 


formula    (5)    page    19   ij   /?  =  —-!  jr,^)  an<^    w^  ^e  ^e   same 


whether  the  back  is  vertical  or  inclined. 

58.  Approximate  Base  Width.  With  regard  to  the  drop  wall 
itself,  owing  to  the  overfall  of  water  and  possible  impact  of  floating 
timber,  ice,  or  other  heavy  bodies,  a  wide  crest  is  a  necessity.  A 
further  strengthening  is  effected  by  adopting  the  trapezoidal  profile. 
The  ordinary  approximate  rule  for  the  base  width  of  a  trapezoidal 
weir  wall  will  be  either 

(H+d) 


VP 
or 

&=—  (16a) 

Vp 

The  correctness  of  either  will  depend  on  various  considerations, 
such  as  the  value  of  d,  the  depth  of  the  overfall,  that  of  hi  or  velocity 
head  and  also  of  /),  the  depth  of  the  tail  water;  the  inclination  given 
to  the  back,  and  lastly,  whether  the  weir  wall  is  founded  on  a  porous 
material  and  is  consequently  subject  to  loss  of  weight  from  uplift. 
Hence  the  above  formulas  may  be  considered  as  approximate  only 
and  the  base  width  thus  obtained  subject  to  correction,  which  is 
easiest  studied  by  the  graphical  process  of  drawing  the  resultant  on 
to  the  base,  ascertaining  its  position  relative  to  the  middle  third 
boundary. 

59.    Approximate  Crest  Width.    With  reference  to  crest  width, 
it  may  be  considered  to  vary  from 

k  =  ^JJT+d  (17) 

to 

(18) 


the  former  gives  a  width  sufficient  for  canal,  or  reservoir  waste  weir 
walls,  but  the  latter  is  more  suitable  for  river  weirs,  and  is  quite  so 
when  the  weir  wall  is  submerged  or  drowned. 

In  many  cases,  however,  the  necessity  of  providing  space  for 
falling  shutters  or  for  cross  traffic  during  times  when  the  weir  is  not 


78  DAMS  AND  WEIRS 

acting,  renders  obligatory  the  provision  of  an  even  wider  crest. 
With  a  moderate  width,  a  trapezoidal  outline  has  to  be  adopted,  in 
order  to  give  the  requisite  stability  to  the  section.  This  is  formed 
by  joining  the  edge  of  the  crest  to  the  toe  of  the  base  by  a  straight 

line,  the  base  width  of  -     -      being  adopted,  as  shown  in  Fig.  45. 

Vp 

When  the  crest  width  exceeds  the  dimensions  given  in  formula 
(18),  the  face  should  drop  vertically  till  it  meets  the  hypothenuse 
of  the  elementary  profile,  as  is  the  case  with  the  pentagonal  profile 
of  dams.  An  example  of  this  is  given  in  Fig.  52  of  the  Dhukwa 
weir.  The  tentative  section  thus  outlined  should  be  tested  by 
graphical  process  and  if  necessary  the  base  width  altered  to  conform 
with  the  theory  of  the  middle  third. 

In  Fig.  45  is  given  a  diagram  of  a  trapezoidal  weir  60  feet  high 
with  d  =  l5  feet.  According  to  formula  (17)  the  crest  width  should 
be  V75  =  8.7  feet,  and  according  to  (18),  7.74+3.87  =  11.6.  An  aver- 
age of  10  feet  has  been  adopted,  which  also  equals  — -  The  profile 

P 

therefore,  exactly  corresponds  with  the  elementary  triangle  canted 
forward  and  truncated  at  the  overfall  crest. 

60.  Graphical  Process.  In  graphical  diagrams,  as  has  already 
been  explained,  wherever  possible  half  widths  of  pressure  areas  are 
taken  off  with  the  compasses  to  form  load  lines,  thus  avoiding  the 
arithmetical  process  of  measuring  and  calculating  the  areas  of  the 
several  trapezoids  or  triangles,  which  is  always  liable  to  error.  There 
are,  however,  in  this  case,  three  areas,  one  of  which,  that  of  the 
reverse  water  pressure,  has  an  altitude  of  only  half  of  the  others. 
This  difficulty  is  overcome  by  dividing  its  half  width  by  2.  If  one 
height  is  not  an  exact  multiple,  as  this  is  of  H,  a  fractional  value 
given  to  the  representing  line  in  the  polygon  will  often  be  found  to 
obviate  the  necessity  of  having  to  revert  to  superficial  measures. 
The  application  of  the  reverse  pressure  PI  here  exhibited  is  similar 
to  that  shown  in  Fig.  16;  it  has  to  be  combined  with  R9  which  latter 
is  obtained  by  the  usual  process.  This  combination  is  effected  in 
the  force  polygon  by  drawing  a  line  PI  equal  to  the  representative 
area,  or  half  width  of  the  back  pressure,  in  a  reverse  direction  to  P. 
The  closing  line  RI  is  then  the  final  resultant.  On  the  profile  itself 
the  force  line  PI  is  continued  through  its  center  of  gravity  till  it 


DAMS  AND  WEIRS  79 

intersects  R,  from  which  point  RI  is  drawn  to  the  base.  If  this  por- 
tion of  the  face  of  the  weir  is  very  flat,  as  is  sometimes  the  case,  Pi 
may  be  so  deflected  as  to  intersect  R  below  the  base  altogether  as 
is  shown  in  Fig.  50.  In  such  event,  RI  is  drawn  upward  instead  of 
downward  to  intersect  the  base.  The  effect  of  PI  is  to  throw  the 
resultant  RI  farther  inward  but  not  to  any  great  extent.  It  im- 
proves the  angular  direction  of  R,  however. 

Reverse  Pressure.  A  dam  is  usually,  but  not  invariably,  exempt 
from  the  effect  of  reverse  pressure.  This  reverse  water  pressure  is 
generally,  as  in  this  case,  favorable  to  the  stability  of  the  weir,  but 
there  are  cases  when  its  action  is  either  too  slight  to  be  of  service  or 
is  even  detrimental.  This  occurs  when  the  face  of  the  weir  wall  is 
much  inclined,  which  points  to  the  equiangular  profile  being  most 
suitable.  An  example  illustrative  of  the  above  remarks  is  given 
later  in  Fig.  50  of  the  Folsam  dam. 

As  the  moments  of  the  horizontal  pressure  of  water  on  either 
side  of  the  weir  wall  vary  almost  with  the  cubes  of  their  height,  it 
is  evident  that  a  comparatively  low  depth  of  tail  water  will  have  but 
small  influence  and  may  well  be  neglected.  When  a  vertical  back 
is  adopted,  the  slope  is  all  given  to  the  face;  by  which  the  normal 
reverse  water  pressure  is  given  a  downward  inclination  that  reduces 
its  capacity  for  helping  the  wall. 

61.  Pressures  Affected  by  Varying  Water  Level.  Calculations 
of  the  depths  of  water  passing  over  the  weir  or  rather  the  height  of 
reservoir  level  above  the  weir  crest,  designated  by  d,  and  of  the 
corresponding  depth  D  in  the  tail  channel,  are  often  necessary  for 
the  purpose  of  ascertaining  what  height  of  water  level  upstream, 
or  value  of  d,  will  produce  the  greatest  effect  on  the  weir  wall.  In 
low  submerged  or  drowned  weirs,  the  highest  flood  level  has  often 
the  least  effect,  as  at  that  time  the  difference  of  levels  above  and 
below  the  weir  are  reduced  to  a  minimum.  This  is  graphically 
shown  in  Fig.  46,  which  represents  a  section  of  the  Narora  dwarf 
weir  wall,  to  which  further  reference  will  be  made  in  section  124, 
Part  II.  In  this  profile  two  resultant  pressures,  R  and  Rit  are 
shown,  of  which  RI,  due  to  much  lower  water  level  of  the  two  stages 
under  comparison,  falls  nearer  the  toe  of  the  base. 

The  Narora  weir,  the  section  of  the  weir  wall  of  which  is  so 
insignificant,  is  built  across  the  Ganges  River  in  Upper  India  at  the 


80 


DAMS  AND  WEIRS 


head  of  the  Lower  Ganges  Canal,  Fig.  93.  The  principal  part  of 
this  work,  which  is  founded  on  the  river  sand,  consists  not  in  the  low 
weir  wall,  although  that  is  f  mile  long,  but  in  the  apron  or  floor, 
which  has  to  be  of  great  width,  in  this  case  200  feet. 

As  will  be  seen  in  Fig.  46  the  flood  level  of  the  Ganges  is  16  feet 
above  the  floor  level,  while  the  afflux,  or  level  of  the  head  water,  is 
two  feet  higher.  The  river  discharges  about  300,000  second-feet 
when  in  flood.  When  full  flood  occurs,  the  weir  is  completely 
drowned,  but  from  the  diagrams  it  will  be  seen  that  the  stress  on 
the  wall  is  less  when  this  occurs  than  when  the  head  water  is 


AFFLUX       1 8-0" 


Fig.  46.     Section  of  Narora  Dwarf  Weir  Wall  across  Ganges  River  in  Upper  India 

much  lower.      This  result  is   due  to  the  reverse  pressure  of  the 
tail  water. 

The  rise  of  the  river  water  produces,  with  regard  to  the  stress 
induced  on  the  weir,  three  principal  situations  or  "stages"  which  are 
enumerated  below. 

(1)  When  the  head  water  is  at  weir  crest  level;  except  in  cases 
where  a  water  cushion  exists,  natural  or  artificial,  the  tail  channel 
is  empty,  and  the  conditions  are  those  of  a  dam. 

(2)  When  the  level  of  the  tail  water  lies  below  weir  crest  level 
but  above  half  the  height  of  the  weir  wall.     In  this  case  the  recip- 
rocal depth  of  the  head  water  above  crest  is  found  by  calculation 


DAMS  AND  WEIRS  81 

(3)  At  highest  flood  level,  the  difference  between  the  head 
and  tail  water  is  at  a  minimum.  In  an  imsubmerged  weir  or  over- 
fall dam  the  greatest  stress  is  generally  produced  during  stage  (3). 
In  a  submerged  weir  the  greatest  stress  is  produced  during  stage  (2). 

62.  Moments  of  Pressure.  The  moments  of  the  horizontal 
water  pressure  on  either  side  of  a  wall  are  related  to  each  other  in 
proportion  to  the  cubes  of  their  respective  depths.  In  cases  where 
the  wall  is  overflowed  by  the  water,  the  triangle  of  pressure  of  the 
latter,  as  we  have  seen,  is  truncated  at  the  weir  crest.  The  moment 
(37)  of  this  trapezoidal  area  of  pressure  will  be  the  product  of  its 
area  with  h,  or  the  product  of  the  expressions  in  formula  (1)  and 
formula  (5)  as  follows: 


or 

M=^  (H+3d)  (19) 

That  of  the  opposing  tail  water  will  be  M  =  —  —  ,  the  difference  of 

6p 

these  two  being  the  resultant  moment.     For  example,  in  the  case 
shown  in  Fig.  46,  during  stage  (1)  77  =  10,  Z)  =  0,  unbalanced  moment 


.  In  stage  (2)  77=10,^  =  3.5,  and  D  =  10.    Then 

bp  p 

the  unbalanced  moment  will  be  ~  [(100  X  20.5)  -  1000]  =175—  . 

6p  p 

In  stage  (3)  77=10,  D  =  16;  d  =  8,  and  D-  77  =  6  feet.  There  will 
thus  be  two  opposing  trapezoids  of  pressure,  and  the  difference  in 
their  moments  will  be 

^(100X34)     ti>(100X28)       ^w 
-  -  ---  -  -  =  100  — 

op  6p  p 

Thus  stage  (2)  produces  the  greatest  effect,  the  least  being  stage 
(3).  In  this  expression  (w)  symbolizes,  as  before,  the  unit  weight 
of  water,  per  cubic  foot  or,  rfa  ton. 

In  spite  of  this  obvious  fact,  many  weir  wall  sections  have  been 
designed  under  the  erroneous  supposition  that  the  overturning 
moment  is  greatest  when  the  upper  water  is  at  crest  level  and  the  tail 
channel  empty,  i.e.,  at  a  time  when  the  difference  of  levels  above  and 


82 


DAMS  AND  WEIRS 


HOlJ.  V 10 


S£3U9tJI    JO  J-NIOd 


below  the  weir  is  at  a 
maximum,  or  at  full 
flood  when  the  differ- 
ence is  at  a  minimum. 
63.  Method  of 
Calculating  Depth  of 
Overfall.  During  the 
second  stage  of  the 
river  the  value  of  d, 
the  depth  of  the  over- 
fall, will  have  to  be 
calculated.  To  effect 
this  the  discharge  of 
the  river  must  first  be 
estimated  when  the 
surface  reaches  the 
crest  level  of  the  weir, 
which  is  done  by  use 
of  the  formula,  Q  = 
Ac^rSj  given  in  sec- 
tion 35,  page  47  of 
"Hydraulics,  Ameri- 
can School  of  Corre- 
spondence", A  being 
the  area,  equal  to  d 
times  length  of  weir 
(c)  Kutter's  coeffi- 
cient, (r)  the  hydraulic 
mean  radius,  and  s, 
the  surface  grade  or 
slope  of  the  river.  The 
discharge  for  the  whole 
river  should  now  be 
divided  by  the  length 
of  the  weir  crest,  the 
quotient  giving  the 
unit  discharge,  or  that 
per  foot  run  of  the  weir. 


DAMS  AND  WEIRS  83 

The  depth  required  to  pass  this  discharge  with  a  free  overfall 
is  found  by  use  of  Francis'  formula  of  3.33dJ  or  a  modification  of  it 
for  wide  crest  weirs  for  which  tables  are  most  useful.  See  "Hydrau- 
lics", section  24,  p.  30. 

For  example,  supposing  the  river  discharge  with  tail  water 
up  to  crest  level  is  20  second-feet  per  foot  run  of  the  weir.  Then 
3.33d1  =  20.  Whence  d*  =  6  and  d=3&  =  3.3  feet.  This  ignores 
velocity  of  approach,  a  rough  allowance  for  which  would  be  to 
decrease  d  by  (hi)  the  velocity  head,  or  by  .0155  F2. 

64.  Illustrative  Example.    Fig.  47  illustrates  an  'assumed  case. 
Here  the  weir  is  15  feet  high,  3  stages  are  shown: 

(1)  When  head  water  is  at  crest  level; 

(2)  When  tail  water  is  7|  feet  deep,  and  the  reciprocal  depth  of  the 

head  water  is  assumed  as  4  feet ;  and 

(3)  \Vith  tail  water  at  crest  level  and  head  water  assumed  7  feet 

deep  above  crest. 

The  three  resultants  have  been  worked  out  graphically.  From 
their  location  on  the  base  the  greatest  stress  is  due  to  7?2,  i.e., 
stage  (2). 

The  hydraulic  gradients  of  all  three  stages  have  been  shown  with 
an  assumed  rear  and  fore  apron  on  floor.  In  (1)  more  than  half  the 
weir  body  lies  below  the  piezometric  line,  which  here  corresponds 
with  the  hydraulic  gradient,  wiiile  in  (2)  nearly  the  whole  lies  below 
this  line  and  in  (3)  entirely  so. 

Owing  to  this  uplift  it  is  well  always  to  assume  the  s.g.  of  a  weir 
wall  under  these  conditions  as  reduced  by  immersion  to  a  value  of 
p  —  1 .  In  these  cases  the  triangles  of  wrater  pressure  are  shown  with 

their  bases  made ,  or  — ,  instead  of  — .  Actually,  however,  the 

resistance  of  the  weir  wall  to  overturning  relative  to  its  base  at  floor 
level  is  not  impaired  by  flotation,  but  as  weight  in  these  cases  is  a 
desideratum,  the  weir  wall  should  be  designed  as  if  this  were  the  case. 
The  rear  apron  is  evidently  subject  to  no  uplift,  but  the  fore  apron  is, 
and  its  resisting  power,  i.e.,  effective  weight,  is  impaired  by  flotation. 
See  section  52  and  also  the  later  sections  on  "  Submerged  Weirs  in 
Sand",  Part  II. 

65.  Examples  of  Existing  Weirs.    Some  examples  of  existing 
weirs  will  now  be  given.    Fig.  48  is  a  profile  of  the  LaGrange  over- 


84 


DAMS  AND  WEIRS 


fall  dam  at  the  head  of  the  Modesto  and  Tuolumne  canals,  Fig.  49. 
No  less  than  13  feet  depth  of  water  passes  over  its  crest,  2  feet  being 


El.  315 


Fig.  48.     Profile  of  LaGrange  Overfall  Dam  at  Head  of  Modesto  and  Tuolumne  Canals 


\CAtiAL  HEAD 


added  to  allow  for  velocity  of  ap- 
proach. It  is  built  on  a  curve  of 
300  feet  radius.  The  graphical 
analysis  of  the  section  shows  that 
the  resultants  (R.E.)  and  (R.F.) 
drawn  on  the  profile  fall  within 
the  middle  third.  In  this  process 
the  reverse  pressure  due  to  tail 
water  has  been  neglected.  Its 
effect  will  be  small. 

It  is  a  doubtful  point  whether 
the  reverse  pressure  actually  exer- 
cised is  that  due  to  the  full  depth 
of  the  tail  water.  The  overflow 
causes  a  disturbance  and  probably  more  or  less  of  a  vacuum  at 
the  toe  of  the  weir  wall,  besides  which  the  velocity  of  impact  causes 
a  hollow  to  be  formed  which  must  reduce  the  reverse  pressure.  In 
some  instances,  as  in  the  case  of  the  Granite  Reef  dam,  Fig.  55,  the 


Fig.  49.     Location  Plan  of  LaGrange  Weir 


DAMS  AND  WEIRS  85 

effective  deptb  of  the  tail  water  is  assumed  as  only  equal  to  that  of 
the  film  of  overflow.  This  appears  an  exaggerated  view.  How- 
ever, in  a  aigh  overfall  dam,  the  effect  of  the  reverse  is  often  so  small 
that  it  can  well  be  neglected  altogether.  In  cases  where  the  tail 
water  rises  to  f  or  more  of  the  height  of  the  dam  its  effect  begins  to 
be  considerable,  and  should  be  taken  into  account. 

66.  Objections   to   "Ogee"    Overfalls.    Professional   opinion 
seems  now  to  be  veering  round  in  opposition  to  the  "bucket"  or 
curved  base  of  the  fore  slope  which  is  so  pronounced  a  feature  in 
American  overfall  dams.    Its  effects  are  undoubtedly  mischievous, 
as  the  destructive  velocity  of  the  falling  water  instead  of  being 
reduced  as  would  be  the  case  if  it  fell  direct  into  a  cushion  of  water, 
is  conserved  by  the  smooth  curved  surface  of  the  bucket.    In  the 
lately  constructed  Bassano  hollow  dam  (see  Figs.  84  and  85,  Part  II), 
the  action  of  the  bucket  is  sought  to  be  nullified  by  the  subsequent 
addition  of  baffles  composed  of  rectangular  masses  of  concrete  fixed 
on  the  curved  slope.    The  following  remarks  in  support  of  this 
view  are  excerpted  from  "The  Principles  of  Irrigation  Engineering" 
by  Mr.  F.  H.  Newell,  formerly  Director  of  United  States  Reclamation 
Service.     "Because  of  the  difficulties  involved  by  the  standing  wave 
or  whirlpool  at  the  lower  toe  of  overflow  dams,  this  type  has  been 
made  in  many  cases  to  depart  from  the  conventional  curve  and  to 
drop  the  water  more  nearly  vertically  rather  than  to  attempt  to 
shoot  it  away  from  the  dam  in  horizontal  lines." 

67.  Folsam  Weir.    Fig.  50  is  of  the  Folsam  weir  at  the  head 
of  the  canal  of  that  name.     It  is  remarkable  for  the  great  depth  of 
flood  water  passing  over  the  crest  which  is  stated  to  be  over  30  feet 
deep.    The  stress  lines  have  been  put  on  the  profile  with  the  object 
of  proving  that  the  reverse  pressure  of  the  water,  although  nearly 
40  feet  deep  has  a  very  small  effect.    This  is  due  to  the  flat  inclina- 
tion given  to  the  lower  part  of  the  weir,  which  has  the  effect  of 
adding  a  great  weight  of  water  on  the  toe  where  it  is  least  wanted 
and  thus  the  salutary  effect  of  the  reverse  pressure  is  more  than 
neutralized.    The  section  is  not  too  heavy  for  requirements,  but  econ- 
omy would  undoubtedly  result  if  it  were  canted  forward  to  a  nearly 
equiangular  profile,  and  this  applies  to  all  weirs  having  deep  tail  water, 
and  to  drowned  weirs.     It  will  be  noted  that  a  wide  crest  allows 
but  very  little  consequent  reduction  in  the  base  width  in  any  case. 


86 


DAMS  AND  WEIRS 


The  stress  diagram  in  Figs.  50  and  50a  are  interesting  as  show- 
ing the  method  of  combining  the  reverse  pressures  with  the  ordinary 
Haessler's  diagram  of  the  direct  water  pressure.  The  profile  is 
divided  into  three  parts  as  well  as  the  direct  water  pressure,  whereas 
the  reverse  pressure  which  only  extends  for  the  two  lower  divisions 
is  in  two  parts.  The  stress  diagrams  present  no  novel  features  till 
Rz  is  reached.  This  force  on  the  profile  comes  in  contact  with  reverse 
force  1"  before  it  reaches  its  objective  31.  The  effect  of  the  reverse 


HEAD  WATER    H '+  d 

~  T\ 


Fig.  50.     Graphical  Analysis  of  Folsam  Weir 

pressure  is  to  deflect  the  direction  of  the  resultant  in  the  direction 
of  /?3,  which  latter,  as  shown  in  the  force  polygon,  Fig.  50a,  is  the 
resultant  of  1",  set  out  from  the  point  b,  and  of  #2-  The  new  result- 
ant J?s  continues  till  it  meets  31.  The  resultant  of  R2  and  31  is  the 
reverse  line  drawn  upward  to  meet  the  vertical  force  3,  parallel  to 
its  reciprocal  in  Fig.  50a,  which  is  the  dotted  line  joining  the  termi- 
nation of  31,  i.e.,  (a)  with  that  of  1". 

Following  the  same  method  the  resultant  7?4  is  next  drawn 
downward  to  meet  2",  which  latter  in  the  force  polygon  is  set  out 


DAMS  AND  WEIRS  87 

from  the  ternination  of  the  vertical  (3).  The  resultant  of  7?4  and 
2"  is  the  final  R5.  This  line  is  drawn  upward  on  the  profile  inter- 
secting the  base  at  B.  If  the  reverse  pressure  were  left  out  of  con- 
sideration, the  force  R%  would  continue  on  to  its  intersection  with 
31  and  thence  the  reverse  recovery  line  drawn  to  meet  (3)  will  be 
parallel  to  ba  (not  drawn)  in  the  force  polygon.  This  reverse  line 
will  intersect  the  line  (3)  in  the  profile  almost  at  the  same  spot 
as  before. 

The  final  line  will  be  parallel  to  its  reciprocal  ca  (not  drawn 
in  Fig.  50a)  and  will  cut  the  base  outside  the  intersection  of  R$.  To 
prevent  confusion  these  lines  have  not  been  drawn  on;  this  proves 
that  the  effect  of  the  reverse  pressure  is  detrimental  to  the  stability 
of  the  wrall,  except  in  the  matter  of  the  inclination  of  R$.  If  the 
profile  were  tilted  forward  this  would  not  be  so.  If  PI  the  resultant 
water  pressure  at  the  rear  of  the  wall  be  drawn  through  the  profile 
to  intersect  the  resultant  of  all  the  vertical  forces,  viz,  1+2+3+?;1 
+fl2,  this  point  will  be  found  to  be  the  same  as  that  obtained  by 
producing  the  final  R$  backwards  to  meet  PI. 

Determination  of  PI.  To  effect  this,  the  position  of  PI  has  to 
be  found  by  the  following  procedure:  The  load  line  db,  Fig.  50a,  is 
continued  to  /,  so  as  to  include  the  forces  3,  Vi,  and  v2.  The  rays 
oc,  oj,  and  ol  are  drawn;  thus  a  new  force  polygon  dol  is  formed  to 
which  the  funicular,  Fig.  50b,  is  made  reciprocal.  This  decides  the 
position  of  W,  or  of  1+2+3,  viz,  the  center  of  pressure  (R.  E.)  as 
also  that  of  W-\-Vi+v2  which  latter  are  the  reverse  pressure  loads. 
The  location  of  PI  is  found  by  means  of  another  funicular  polygon  C 
derived  from  the  force  polygon  oad,  by  drawing  the  rays  oa,  of,  and 
oe;  PI  is  then  drawn  through  the  profile  intersecting  the  vertical 
resultant  last  mentioned  at  A.  The  line  AB  is  then  coincident 
with  R$  on  Fig.  50.  The  vertical  line  through  A  is  not  N,  i.e.,  is 
not  identical  with  the  vertical  in  Fig.  50,  for  the  reason  that  N  is 
the  resultant  of  all  the  vertical  forces,  whereas  the  vertical  in  ques- 
tion is  the  centroid  of  pressure  of  all  the  vertical  force  less  Wi,  the 
weight  of  water  overlying  the  rear  slope  of  the  wall.  The  location 
of  N  is  found  by  drawing  a  horizontal  P  through  the  intersection 
of  PI  with  a  line  drawn  through  the  e.g.  of  the  triangle  of  water 
pressure  w,  this  will  intersect  the  back  continuation  of  BA  at  c. 
A  vertical  CD  through  this  point  will  correspond  with  that  marked 


88 


DAMS  AND  WEIRS 


N  in  Fig.  50.  The  profile,  Fig.  51,  is  a  reproduction  of  that  shown 
Fig.  50  in  order  to  illustrate  the  analytical  method  of  calculation  or 
that  by  moments. 

68.  Analytical  Method.  The  incidence  of  the  resultant  R  is 
required  to  be  as  ascertained  on  two  bases,  one  the  final  base  and 
the  other  at  a  level  13  feet  higher.  The  section  of  the  wall  as  before, 
is  divided  into  three  parts:  (1)  of  area  840  square  feet,  (2)  of  1092,  and 
(3)  of  838  square  feet.  The  position  of  the  e.g.  of  (1)  is  found  by  for- 
mula (7)  to  be  15.15  feet  distant  from  a  the  heel  of  the  base  and  will  be 


1747      I 


FLOOD     LEVEL 


^V~      ^V       ' 

-«i        PV ' 

^/f.//////»///////////f//////V///////////^>>^    T 


Fig.  51.    Diagram  of  Folsam  Weir  Illustrating  Analytical  Method  of  Calculation 

15.65  feet  from  b.  That  of  (2)  is  32.3  feet  distant  from  its  heel  6.  Th* 
reduced  area  of  the  water  overlying  the  back  down  to  b  is  estimated 
at  26  square  feet  and  by  formula  (6)  to  be  .5  feet  distant  from  b. 
Again  the  reduced  area  of  the  reverse  water  overlying  the  fore 

18  5 
slope  Vi  is  92  square  feet  and  the  distance  of  its  e.g.  from  b  is  55 — 

*, 

=  48.8  feet.    The  moments  of  all  these  vertical  forces  equated 

with  that  of  their  sum  (N)  about  the  point  b  will  give  the  position 
of  N  relative  to  b. 


DAMS  AND  WEIRS  89 

Thus 

(1)  840X15.65  =  13146 

(2)  1092X23.3  =25443 
(w)      26  X     .5  =       13 
(t>i)      92X48.8  =  4490 

2050  X  x       =  43092  =  Moment  of  N 
x    =       21  feet,  nearly 

To  obtain  the  distance  /  between    N  and  R,  f=     p    —  ?'.     Now 

the  reduced  area  of  P  =  1257  and  the  height  of  the  e.g.  of  the 
trapezoid  having  its  base  at  b,  and  its  crest  level  with  that  of  the 
wall  is  calculated  by  formula  (6),  to  be  22.1  feet.  Again  the  reduced 
area  of  the  reverse  water  pressure  triangle  pi  is  120  square  feet, 
the  height  of  its  e.g.  above  base  is  8  feet.  Consequently: 

,_  (1257X22.1)  -  (120X8)  _  26820 
2050  ~  2050  = 

For  the  lower  base,  the  statement  of  moments  about  c  is  as  fol- 
lows, i>2  being  240,  and  its  distance  65  feet  by  formula  (6)  . 

(  N)  2050  X  (21  +  .3)  =43665 

(3)   838X32.3        =27067 

(w)      10X     .15 

fa)    240X65 

Total       3138  Xz 

86334      ,_ 


Now  /i  =  MPI  ~pip*  }  fr  being  tne  distance  between  NI  and 

NI 

The  value  of  Pi,  the  trapezoid  of  water  pressure  down  to  the  base 
c,  is  1747  square  feet  and  the  height  of  its  e.g.  by  formula  (19)  or 

(5)  is  27  feet,  that  of  (pi-\-pz)  is  285  square  feet  and  its  lever  arm 
07 

^  =  12Jfeet.    Then 
o 

(1747X27)-(285X12^47169-3514^_43655  f 

3138  3138  3138 

The  positions  of  N  and  NI  being  obtained,  the  directions  of 
R  and  RI  are  lines  drawn  to  the  intersections  of  the  two  verticals 


90  DAMS  AND  WEIRS 

N  and  NI  with  two  lines  drawn  through  the  c.g.'s  of  the  trapezoid  of 
pressure  reduced  by  the  moment  of  the  reverse  pressure,  if  any,  or  by 
(P—p).  This  area  will  consist,  as  shown  in  the  diagram,  of  a  trape- 
zoid superposed  on  a  rectangle;  by  using  formula  (5)  section  1,  the 
positions  of  the  e.g.  of  the  upper  trapezoid  is  found  to  be  12.58  feet 
above  the  base  at  a,  while  that  of  the  lower  is  at  half  the  depth  of 
the  rectangle,  then  by  taking  moments  of  these  areas  about  6,  the 
height  of  the  e.g.  is  found  to  be  23.6  feet  above  the  base  at  b,  while 
the  height  for  the  larger  area  [Pi—(pi+pz)]  down,  to  c  is  27  feet. 

In  the  graphical  diagram  of  Fig.  51a  the  same  result  would  be 
obtained  by  reducing  the  direct  pressure  by  the  reverse  pressure 
area.  Thus  in  the  force  diagram  the  vertical  load  line  would  remain 
unchanged  but  the  water-pressure  load  line  would  be  shorter  being 
P—p  and  Pi—  (PI+PZ),  respectively.  This  would  clearly  make  no 
difference  in  the  direction  of  the  resultants  R  and  RI  and  would  save 
the  two  calculations  for  the  c.g.'s  of  P  and  PI. 

This  weir  is  provided  with  a  crest  shutter  in  one  piece,  150  feet 
long,  which  is  raised  and  lowered  by  hydraulic  jacks  chambered 
in  the  masonry  of  the  crest  so  that  they  are  covered  up  by  the  gate 
when  it  falls.  This  is  an  excellent  arrangement  and  could  be  imi- 
tated with  advantage.  The  shutter  is  5  feet  deep.  The  width 

TT     i        7 

at  base  of  lamina  2  of  this  weir  is  55  feet,  or  very  nearly  — -— , 

VP 

formula  (16). 

69.  Dhukwa  Weir.  A  very  similar  work  is  the  Dhukwa  weir 
in  India,  Fig.  52,  which  has  been  recently  completed. 

This  overfall  dam  is  of  pentagonal  section.  Owing  to  the  width 
of  the  crest  this  is  obviously  the  best  outline. 

The  stress  resultant  lines  have  been  drawn  on  the  profile,  which 
prove  the  correctness  of  the  base  width  adopted.  The  tail  water 
does  not  rise  up  to  half  the  height  of  the  weir.  Consequently  the 

formula —  is  applicable  in  stage  3.    The  effect  of  the  tail  water 

Vp 

is  practically  nil.    According  to  this  formula  the  base  width  would 

2 
be  63  X —  =  42  feet,  which  it  almost  exactly  measures — a  further 

o 

demonstration  of  the  correctness  of  the  formula. The  crest  width 

should  be,  according  to   formula  (18),  V50+Vl3  =  ll   feet.    The 


DAMS  AND  WEIRS 


width  of  17  feet  adopted  is  necessary  for  the  space  required  to  work 
the  collapsible  gates.  These  are  of  steel,  are  held  in  position  by 
struts  connected  with  triggers,  and  can  be  released  in  batches  by 
chains  worked  from  each  end.  The  gates,  8  feet  high,  are  only  10 
feet  wide.  This  involves  the  raising  and  lowering  of  400  gates,  the 
weir  crest  being  4000  feet  long.  The  arrangement  adopted  in  the 
Folsam  weir  of  hydraulic  jacks  operating  long  gates  is  far  superior. 
An  excellent  feature  in  this  design  is  the  subway  with  occasional  side 
chambers  and  lighted  by  openings,  the  outlook  of  which  is  under- 
neath the  waterfall,  and  has  the  advantage  of  relieving  any  vacuum 
under  the  falling  water. 


AFFLUX  -r  h 


El.  903 


ftDDD 


Fig.  52.     Graphical  Analysis  of  Profile  of  Dhukwa  Weir  in  India 

The  subway  could  be  utilized  for  pressure  pipes  and  for  cross 
communication,  and  the  system  would  be  most  useful  in  cases  where 
the  obstruction  of  the  crest  by  piers  is  inadvisable.  The  weir  is 
4000  feet  long  and  passes  800,000  second-feet,  with  a  depth  of  13 
feet.  The  discharge  is,  therefore,  200  second-feet  per  foot  run"  of 

200 
weir,  which  is  very  high.     The  velocity  of  the  film  will  be  —  =  15.4 

lo 

feet  per  second.  With  a  depth  of  13  feet  still  water,  the  discharge 
will  be  by  Francis'  formula,  156  second-feet  per  foot  run.  To  produce 
a  discharge  of  200  feet  per  second,  the  velocity  of  approach  must  be 
about  10  feet  per  second.  This  will  add  2.3  feet  to  the  actual  value 


92 


DAMS  AND  WEIRS 


of  d,  raising  it  from  13  to  15.3  feet  which  strictly  should  have  been 
done  in  Fig.  52. 

70.  Mariquina  Weir.  Another  high  weir  of  American  design, 
Fig.  53,  is  the  Mariquina  weir  in  the  Philippines.  It  has  the  ogee 
curve  more  accentuated  than  in  the  LaGrange  weir.  The  stress 
lines  have  been  drawn  in,  neglecting  the  effect  of  the  tail  water 
which  will  be  but  detrimental.  The  section  is  deemed  too  heavy 
at  the  upper  part  and  would  also  bear  canting  forward  with  advan- 
tage, but  there  are  probably  good  reasons  why  an  exceptionally  solid 


%7>^Z  WATERj4\SEL  (ASSUMED) 


Fig.  53.     Profile  of  Mariquina  Weir  in  the  Philippines 

crest  was  adopted.  The  ogee  curve  also  is  a  matter  on  which 
opinion  has  already  been  expressed. 

71.  Granite  Reef  Weir.  The  Granite  Reef  weir  over  the 
Salt  River,  in  Arizona,  Figs.  54  and  55,  is  a  work  subsidiary  to  the 
great  Roosevelt  dam  of  which  mention  was  previously  made. 

It  is  founded  partly  on  rock  and  partly  on  boulders  and  sand 
overlying  rock.  The  superstructure  above  the  floor  level  is  the 
same  throughout,  but  the  foundations  on  shallow  rock  are  remark- 
able as  being  founded  not  on  the  rock  itself,  but  on  an  interposed 
cushion  of  sand.  (See  Fig.  54.)  Reinforced  concrete  piers,  spaced 
20  feet  apart,  were  built  on  the  bedrock  to  a  certain  height,  to  clear 


DAMS  AND  WEIRS 


93 


5 


si'..:;/:  •'..'••  SAND  '•••••::y'l^y' 


all  inequalities;  these  were  connected  by  thin  reinforced  concrete 
side  walls;  the  series  of  boxes  thus  formed  were  then  filled  level 
with  sand,  and  the  dam  built  thereon.  This  work  was  completed 
in  1908.  The  portion  of  the  profile  below  the  floor  is  conjectural. 
This  construction  appears  to  be  a  bold  and  commendable  novelty. 
Sand  in  a  confined  space  is  incompressible,  and  there  is  no  reason 
why  it  should  not  be  in  like  situations.  A  suggested  improvement 
would  be  to  abandon  the  piers  and  form  the  substructure  of  two 
long  outer  walls  only,  braced  together  with  rods  or  old  rails  encased 
in  concrete.  Fig.  55  is  the  profile  on  a  boulder  bed  with  rock  below. 

72.  Hydraulic  Condi= 
tions.  The  levels  of  the 
afflux  flood  of  this  river  are 
obtainable  so  that  the  stresses 
can  be  worked  out.  In  most 
cases  these  necessary  statis- 
tics are  wanting.  The  flood 
downstream  has  been  given 
the  same  depth,  12  feet,  as 
that  of  the  film  passing  over 
the  crest.  This  is  clearly 
erroneous.  The  velocity  of 
the  film  allowing  for  5  feet 
per  second  approach,  is  quite 
12  feet  per  second,  that  in 
the  river  channel  could  not 
be  much  over  5  feet,  conse- 

12X 12 

quently  it  would  require  a  depth  of =  28  feet.     The  dam 

o 

would  thus  be  quite  submerged,  which  would  greatly  reduce  the 
stress.  As  previously  stated,  the  state  of  maximum  stress  would 
probably  occur  when  about  half  the  depth  of  flood  passes  over 
the  crest.  However,  the  graphical  work  to  find  the  incidence 
of  the  resultant  pressure  on  the  base  will  be  made  dependent  on  the 
given  downstream  flood  level.  After  the  explanations  already 
given,  no  special  comment  is  called  for  except  with  regard  to  the 
reverse  water  pressure.  Here  the  curved  face  of  the  dam  is  altered 
into  2  straight  lines  and  the  water  pressure  consists  of  two  forces 


Fig.    54.     Section    of    Granite    Reef    Weir 
Showing  Sand  Cushion  Foundations 


94 


DAMS  AND  WEIRS 


having  areas  of  17  and  40,  respectively,  which  act  through  their 
c.g.'s.     Instead  of  combining  each  force  separately  with  the  result- 


ant  (/?)  it  is  more  convenient  to  find  their  resultant  and  combine 
that  single  force  with  (R.)  This  resultant  PI  must  pass  through 
the  intersection  of  its  two  components,  thus  if  their  force  lines  are 


DAMS  AND  WEIRS  95 

run  out  backward  till  they  intersect,  a  point  in  the  direction  of  PI 
is  found.  PI  is  then  drawn  parallel  to  its  reciprocal  in  the  force 
polygon  which  is  also  shown  on  a  larger  scale  at  the  left  of  the  pro- 
file. The  final  resultant  is  RI  which  falls  just  within  the  middle 
third  of  the  base.  Rz  is  the  resultant  supposing  the  water  to  be 
at  crest  level  only.  The  water  in  the  river  is  supposed  to  have 
mud  in  solution  with  its  s.g.  1.4.  The  base  length  of  the  triangle 


*  -n    +u       u  l)     32X1.4 

of  water  pressure  will   then  be  -  -=  —  —  —  =  18.66. 

The  other  water-pressure  areas  are  similarly  treated.  If  the  rear 
curtain  reaches  rock  the  dam  should  not  be  subject  to  uplift.  It 
could,  however,  withstand  sub-percolation,  as  the  hearth  of  riprap 
and  boulders  will  practically  form  a  filter,  the  material  of  the  river 
bed  being  too  large  to  be  disintegrated  and  carried  up  between  the 
interstices  of  the  book  blocks.  The  effective  length  of  travel  would 
then  be  107  feet;  add  vertical  52  feet,  total  159  feet,  H  being  20  feet, 

jf  works  out  to  —  -  =  8  which  ratio  is  a  liberal  allowance  for  a  boulder 
H  20 

bed.  The  fore  curtain  is  wisely  provided  with  weep  holes  to  release 
any  hydrostatic  pressure  that  might  otherwise  exist  underneath  the 
dam.  The  Granite  Reef  dam  has  a  hearth,  or  fore  apron  of  about 
80  feet  in  width.  A  good  empirical  rule  for  the  least  width  for  a 
solid  or  open  work  masonry  fore  apron  is  the  following  : 

L  =  2H+d  (20) 

in  which  //  is  the  height  of  the  permanent  weir  crest  above  floor, 
and  d  is  the  depth  of  flood  over  crest.  In  this  case  #  =  20,  d  =  12; 
least  width  of  floor  should  then  be  40+12  =  52  feet.  The  Bassano 
dam  is  40  feet  high  with  14  feet  flood  over  crest,  the  width  of  hearth 
according  to  this  formula  should  be  94  feet,  its  actual  width  is  80 
feet  which  is  admittedly  insufficient.  With  a  low  submerged  weir, 
formula  (34)  ,  Part  II,  viz,  L  =  3  Vc  H,  will  apply.  Beyond  the  hearth 
a  talus  of  riprap  will  generally  be  required,  for  which  no  rule  can 
well  be  laid  down. 

73.  Nira  Weir.  Fig.  56  is  of  the  Nira  weir,  an  Indian  work. 
Considering  the  great  depth  of  the  flood  waterdown  stream,  the  pro- 
vision of  so  high  a  subsidiary  weir  is  deemed  unnecessary,  a  water 
cushion  of  10  feet  being  ample,  as  floor  is  bed  rock.  The  section  of 


96 


DAMS  AND  WEIRS 


the  weir  wall  itself,  is  considered  to  be  somewhat  deficient  in  base 
width.  Roughly  judging,  the  value  of  H-\-d,  on  which  the  base 
width  is  calculated,  should  include  about  3  or  4  feet  above  crest 
level.  This  value  of  d,  it  is  believed  would  about  represent  the  height 
of  head  water,  which  would  have  the  greatest  effect  on  the  weir. 
The  exact  value  of  d  could  only  be  estimated  on  a  knowledge  of  the 
bed  slope  or  surface  grade  of  the  tail  channel.  The  above  estimate 

would  make  (#-M)=36  feet,  and  with  p  =  2j,  — ^-=24  feet. 

The  top  width,  8.3,  is  just  V//+Vr/,  in  accordance  with  the 
rule  given  in  formula  (18). 

A  section  on  these  lines  is  shown  dotted  on  the  profile.  The 
provision  of  an  8-foot  top  width  for  the  subsidiary  weir  is  quite 


Fig.  56.     Section  of  Nira  Weir  in  India  Showing  Use  of  Secondary  Weir 

indefensible,  while  the  base  width  is  made  nearly  equal  to  the 
height,  which  is  also  excessive.  For  purposes  of  instruction  in  the 
principles  of  design,  no  medium  is  so  good  as  the  exhibition  of  plans 
of  actual  works  combined  with  a  critical  view  of  their  excellencies 
or  defects.  The  former  is  obtainable  from  record  plans  in  many 
technical  works,  but  the  latter  is  almost  entirely  wanting.  Thus 
an  inexperienced  reader  has  no  means  of  forming  a  just  opinion  and 
is  liable  to  blindly  follow  designs  which  may  be  obsolete  in  form 
or  otherwise  open  to  objection. 

74.  Castlewood  Weir.  The  Castlewood  weir,  Fig.  57,  is  of 
remarkable  construction,  being  composed  of  stonework  set  dry, 
enclosed  in  a  casing  of  rubble  masonry.  It  is  doubtful  if  such  a 
section  is  any  less  expensive  than  an  ordinary  gravity  section,  or 


DAMS  AND  WEIRS 


97 


much  less  than  an  arched  buttress  dam  of  type  C.  Shortly  after 
construction,  it  showed  signs  of  failure,  which  was  stated  to  be  due 
to  faulty  connections  with  banks  of  the  river;  but  whatever  the 
cause  it  had  to  be  reinforced,  which  was  effected  by  adding  a  solid 
bank  of  earth  in  the  rear,  as  shown  in  the  figure.  This  involved 
lengthening  the  outlet  pipes.  In  the  overfall  portion  the  bank  must 
have  been  protected  with  riprap  to  prevent  scouring  due  to  the 
velocity  of  the  approach  current. 

75.  American  Dams  on  Pervious  Foundations.  In  the  United 
States  a  very  large  number  of  bulkhead  and  overfall  dams  and  regu- 
lating works,  up  to  over  100  feet  in  height  have  been  built  on  foun- 


Fig.  57.     Section  of  Castlewood  Weir  Showing  Construction  of  Stone  Work  Set  Dry, 
Enclosed  in  Rubble  Masonry 

dations  other  than  rock,  such  as  sand,  boulders,  and  clay.  Most 
of  these,  however,  are  of  the  hollow  reinforced  concrete,  or  scallop 
arch  types,  in  which  a  greater  spread  for  the  base  is  practicable 
than  would  be  the  case  with  a  solid  gravity  dam.  Whenever  a  core 
wall  is  not  run  down  to  impervious  rock,  as  was  the  case  in  the 
Granite  Reef  Overfall  dam,  Fig.  55,  the  matter  of  sub-percolation 
and  uplift  require  consideration,  as  is  set  forth  in  the  sections  on 
"Gravity  Dams"  and  "Submerged  Weirs  on  Sand".  If  a  dam  50 
feet  high  is  on  sand  or  sand  and  boulders,  of  a  quality  demanding 
a  high  percolation  factor  of  say  10  or  12,  it  is  clear  that  a  very  long 
rear  apron  and  deep  rear  piling  will  be  necessary  for  safety. 

All  rivers  bring  down  silt  in  suspension.     When  the  overfall 
dam  is  a  high  one  with  a  crest  more  than  15  or  20  feet  above  river- 


98  DAMS  AND  WEIRS 

bed  level,  the  deposit  that  is  bound  to  take  place  in  rear  of  the 
obstruction  will  not  be  liable  to  be  washed  out  by  the  current,  and 
additional  light  stanching  silt  will  be  deposited  in  the  deep  poo)  of 
comparatively  still  water  that  must  exist  at  the  rear  of  every  high 
dam.  For  a  low  weir  however  this  does  not  follow,  and  i?  deposit  is 
made  it  will  be  of  the  heavier,  coarser  sand  which  is  not  impermeable. 

The  difficulty  and  expense  of  a  long  rear  apron  can  be  sur- 
mounted by  the  simple  expedient  of  constructing  only  a  portion  of 
it  of  artificial  clay,  leaving  the  rest  to  be  deposited  by  the  river 
itself.  To  ensure  safety  the  dam  should  be  constructed  and  reser- 
voir filled,  in  two  or  three  stages,  with  intervals  between  of  sufficient 
length  to  allow  the  natural  deposit  to  take  place.  Thus  only  a  frac- 
tion of  the  protective  apron  need  be  actually  constructed.  Many 
works  are  in  existence  which  owe  their  safety  entirely  to  the  fortu- 
nate but  unrecognized  circumstance  of  natural  deposit  having 
stanched  the  river  bed  in  their  rear,  and  many  failures  that  have 
taken  place  can  only  be  accounted  for  from  want  of  provision  for 
the  safety  of  the  work  against  underneath  scour  or  piping  and  also 
uplift.  The  author  himself  once  had  occasion  to  report  on  the  fail- 
ure of  a  head  irrigation  work  which  was  designed  as  if  on  rock, 
whereas  it  was  on  a  pervious  foundation  of  boulders.  When  it 
failed  the  designers  had  no  idea  of  the  real  cause,  but  put  it  down 
to  a  "treacherous  river",  "ice  move",  anything  but  the  real  reason, 
of  which  they  were  quite  ignorant.  Had  a  rear  apron  of  sufficient 
width  been  constructed,  the  work  would,  be  standing  to  this  day. 

76.  Base  of  Dam  and  Fore  Apron.  The  fore  apron  and  base 
of  an  overfall  dam  or  weir  must  be  of  one  level  throughout  its 
length,  if  the  foundation  is  of  any  other  material  than  rock.  The 
foundation  core  walls  may  have  to  vary  more  or  less  with  the  surface 
of  the  river  bed,  which  is  deep  in  some  places,  and  shallow  in 
others,  but  the  apron  level  should  be  kept  at  or  about  low  water 
level  throughout.  When  a  horizontal  wall  as  an  overfall  dam  is 
built  across  a  river  bed  it  obliterates  the  depressions  and  channels 
in  it,  the  discharge  over  the  weir  is  the  same  at  all  points  or  nearly 
so,  consequently  the  tendency  will  be  to  level  the  bed  downstream  by 
filling  the  hollows  and  denuding  the  higher  parts. 

Under  these  conditions  it  is  evidently  sheer  folly  to  step  up  the 
apron  to  coincide  with  the  section  of  the  river  bed,  as  the  higher 


DAMS  AND  WEIRS 


99 


parts  of  the  bed  are  bound  to  be  in  time  washed  out  by  the  falling 
water  and  deposited  in  the  deeper  channels,  and  portions  of  the 
dam  may  easily  be  undermined.  This  actually  occured  in  one  case. 
77.  Section  of  Spillway  of  St.  Maurice  River  Dam.  Fig.  58  is 
a  section  of  the  spillway  portion  of  the  reinforced  bulkhead  gravity 
dam,  illustrated  in  Fig.  44.  Owing  to  the  absence  of  the  heavy 
crest  of  Fig.  44,  the  back  of  the  spillway  profile  is  provided  with 


24  Tons 
ffeinf. 


Fig.  58.     Diagram  Showing  Profile  of  Spillway  Portion  of  Saint  Maurice  River  Dam 

(See  Fig.  44) 

double  the  amount  of  reinforcement  shown  in  the  former  example. 
One  half,  viz,  1J  inches,  extends  right  down  to  the  base,  while  the 
other  half  stops  short  at  El  280.  This  is  arranged  for  in  the  stress 
diagrams  in  the  same  way  as  explained  in  section  55,  R$  being  the 
final  resultant  on  the  base.  The  line  of  pressure  falls  slightly  out- 
side the  middle  third  in  the  upper  half  of  the  section.  The  effect 
would  be  to  increase  the  tension  in  the  reinforcement  somewhat 
above  the  limit  of  8  tons  per  square  inch.  The  adoption  of  a  trape- 
zoidal profile,  would,  it  is  deemed,  be  an  improvement  in  this  case 
as  well  as  in  the  former. 


I 


DAMS  AND  WEIRS 

PART  II 


ARCHED  DAMS 

78.  General  Characteristics.  In  this  type,  the  whole  dam, 
being  arched  in  plan,  is  supposed  to  be  in  the  statical  condition  of 
an  arch  under  pressure.  As,  however,  the  base  is  immovably  fixed 
to  the  foundations  by  the  frictional  resistance  due  to  the  weight  of 
the  structure,  the  lowest  portion  of  the  dam  cannot  possess  full 
freedom  of  motion  nor  elasticity,  and  consequently  must  act  more  or 
less  as  a  gravity  dam  subject  to  oblique  pressure. 

However  this  may  be,  experience  has  conclusively  proved  that 
if  the  profile  be  designed  on  the  supposition  that  the  whole  is  an 
elastic  arch,  this  conflict  of  stresses  near  the  base  can  be  neglected 
by  the  practical  man.  The  probability  is  that  both  actions  take 
place,  true  arch  action  at  the  crest,  gradually  merging  into  transverse 
stress  near  the  base;  the  result  being  that  the  safety  of  the  dam  is 
enhanced  by  the  combination  of  tangential  and  vertical  stresses  on 
two  planes. 

In  this  type  of  structure,  the  weight  of  the  arch  itself  is  conveyed 
to  the  base,  producing  stress  on  a  horizontal  plane,  while  the  water 
pressure  normal  to  the  extrados,  radial  in  direction,  is  transmitted 
through  the  arch  rings  to  the  abutments.  The  pressure  is,  therefore, 
distributed  along  the  whole  line  of  contact  of  the  dam  with  the  sides 
as  well  as  the  ground.  In  a  gravity  dam,  on  the  other  hand,  the  whole 
pressure  is  concentrated  on  the  horizontal  base. 

Arch  Stress.  The  average  unit  stress  developed  by  the  water 
pressure  is  expressed  by  the  formula 

"Short"  Formula     (21) 
"Short"  Formula    (21a) 


102  DAMS  AND  WEIRS 

in  which  R  is  the  radius  of  the  extrados,  sometimes  measured  to  the 
center  of  the  crest,  //  the  depth  of  the  lamina,  b  its  width,  and  w 
the  unit  weight  of  water  or  ^  ton.  Into  this  formula  p,  the  specific 
gravity  of  the  material  in  the  arch,  does  not  enter.  This  simple 
formula  answers  well  for  all  arched  dams  of  moderate  base  width. 
When,  however,  the  base  width  is  considerable,  as,  say,  in  the  case 
of  the  Pathfinder  dam,  the  use  of  a  longer  formula  giving  the  maxi- 
mum stress  (s)  is  to  be  preferred.  This  formula  is  derived  from  the 
same  principle  affecting  the  relations  of  s  and  Si,  or  of  the  maximum 
and  average  stresses  already  referred  to  in  Part  I  on  "Gravity  Dams". 
The  expression  is  as  follows,  r  being  the  radius  of  the  intrados: 

2R    _RHw      2R 
S\R+r)~     b     XR+r 
or  in  terms  of  R  and  b 


R 
also 


Long"  Formula    (22) 


b  =  R(  1  -  ^h  _££!?  }        "Long"  Formula  (22a) 


79.  Theoretical  and  Practical  Profiles.  In  a  manner  similar 
to  gravity  dams,  the  theoretical  profile  suitable  for  an  arched  dam 
is  a  triangle  having  its  apex  at  the  extreme  water  level,  its  base 
width  being  dependent  on  the  prescribed  limiting  pressure.  Success- 
ful examples  have  proved  that  a  very  high  value  for  s,  the  maximum 
stress,  can  be  adopted  with  safety.  If  it  were  not  for  this,  the  profit- 
able use  of  arched  dams  would  be  restricted  within  the  narrow  limits 
of  a  short  admissible  radius,  as  with  a  low  limit  pressure  the  section 
would  equal  that  of  a  gravity  dam. 

The  practical  profile  is  a  trapezoid,  a  narrow  crest  being  nece# 
sary.  The  water  pressure  area  acting  on  an  arched  dam,  is  naturally 
similar  to  that  in  a  gravity  dam,  the  difference  being,  however, 
that  there  is  no  overturning  moment  when  reverse  pressure  occurs 
as  in  a  weir.  The  difference  or  unbalanced  pressure  acting  at  any 
point  is  simply  the  difference  of  the  direct  and  the  reverse  forces. 
The  areas  of  pressure  on  both  sides,  therefore,  vary  with  the  squares 
of  their  respective  depths. 


DAMS  AND  WEIRS  103 

The  water  pressure  on  an  arch  acts  normally  to  the  surface  of 
its  back  and  is  radial  in  direction;  consequently  the  true  line  of  pres- 
sure in  the  arch  ring  corresponds  with  the  curvature  of  the  arch  and 
has  no  tendency  to  depart  from  this  condition.  There  is,  therefore, 
no  such  tendency  to  rupture  as  is  the  case  in  a  horizontal  circular 
arch  subjected  to  vertical  rather  than  radial  pressure.  This  prop- 
erty conduces  largely  to  the  stability  of  an  arch  under  liquid 
pressure.  This  condition  is  not  strictly  applicable  in  its  entirety 
to  the  case  of  a  segment  of  a  circle  held  rigidly  between  abut- 
ments as  the  arch  is  then  partly  in  the  position  of  a  beam.  The 
complication  of  stress  involved  is,  however,  too  abstruse  for  practical 
consideration. 

80.  Correct  Profile.  As  we  have  already  seen,  the  correct 
profile  of  the  arched  dam  is  a  triangle  modified  into  a  trapezoid 
with  a  narrow  crest.  With  regard  to  arch  stresses,  the  most  favorable 
outline  is  that  with  the  back  of  the  extrados  vertical.  The  reason  for 
this  is  that  the  vertical  stress  due  to  the  weight  of  the  arch,  although 
it  acts  on  a  different  plane  from  the  tangential  stresses  in  the  arch 
ring,  still  has  a  definable  influence  on  the  maximum  induced  stress 
in  the  arch  ring.  The  vertical  pressure  produces  a  transverse  expan- 
sion which  may  be  expressed  as  WxEXm,  in  which  E  is  the  coeffi- 
cient of  elasticity  of  the  material  and  m  that  of  transverse  dilation. 
This  tends,  when  the  extrados  is  vertical,  to  diminish  the  maximum 
stress  in  the  section;  whereas  when  the  intrados  is  vertical  and  the 
back  inclined,  the  modification  of  the  distribution  of  pressure  is 
unfavorable,  the  maximum  stress  being  augmented.  When  the 
trapezoidal  profile  is  equiangular,  an  intermediate  or  neutral  condi- 
tion exists.  A  profile  with  vertical  extrados  should,  therefore,  be 
adopted  whenever  practicable. 

In  very  high  dams,  however,  the  pressure  on  the  horizontal 
plane  of  the  base  due  to  the  weight  of  the  structure,  becomes  so 
great  as  to  even  exceed  that  in  the  arch  ring;  consequently  it  is 
necessary  to  adopt  an  equiangular  profile  in  order  to  bring  the  center 
of  pressure  at,  or  near  to,  the  center  of  the  base,  so  as  to  reduce  the 
ratio  of  maximum  pressure  to  average  pressure  to  a  minimum. 
As  stated  in  the  previous  section,  when  a  vertical  through  the 
center  of  gravity  of  the  profile  passes  through  the  center  of  the  base, 
the  maximum  pressure  equals  the  average,  or  s  =  Si. 


104 


DAMS  AND  WEIRS 


81.  Support    of    Vertical    Water    Loads    in    Arched    Dams. 

When  the  back  of  an  arched  dam  is  inclined,  the  weight  of  the 
water  over  it  is  supported  by  the  base,  the  horizontal  pressure  of 
the  water  alone  acting  on  the  arch  and  being  conveyed  to  the 
abutments.  In  the  case  of  inclined  arch  buttress  dams,  however,  a 
portion  of  the  vertical  load  is  carried  by  the  arch,  increasing  its 
thrust  above  what  is  due  to  the  horizontal  water  pressure  alone. 
This  is  due  to  overhang,  i.  e.,  when  the  c.  g.  falls  outside  the  base. 

82.  Crest  Width.     The  crest  width  of  arched  dams  can  be 
safely  made  much  less  than  that  of  gravity  dams  and  a  rule  of 


\^  Rod.  335' 


k=— 


(23) 


would  seem  to  answer  the  purpose,  unless  rein- 
forcement is  used,  when  it  can  be  made  less. 

EXAMPLES  OF  ARCHED  DAMS 

The  following  actual  examples  of  arched 
dams  will  now  be  given. 

83.  Bear  Valley  Dam.  This  small  work, 
Fig.  59,  is  the  most  remarkable  arched  dam  in 
existence  and  forms  a  valuable  example  of  the 
enormous  theoretical  stresses  which  this  type 
of  vertical  arch  can  stand.  The  mean  radius 
being  335  feet  according  to  formula  (21)  the 
unit  stress  will  be 


RHw 


=  60  tons,  nearly 


Fig.  59.   Section  of  Old  Bear  This  section  would  be  better  if  reversed.    The 

Valley  Dam  ,  .  .  , 

actual  stress  is  probably  half   this   amount. 

This  work  has  now  been  superseded  by  a  new  dam  built  below  it, 
Fig.  77,  section  103. 

84.  Pathfinder  Dam.  This  immense  work,  Fig.  60,  is  built 
to  a  radius  of  150  feet  measured  to  the  center  of  the  crest.  That, 
however,  at  the  extrados  of  the  base  of  the  section  is  186  feet  and  this 
quantity  has  to  be  used  for  the  value  of  R  in  the  long  formula  (22). 
The  unit  stress  then  works  out  to  18  tons,  nearly.  The  actual  stress 
in  the  lowest  arch  ring  is  undoubtedly  much  less,  for  the  reason 


DAMS  AND  WEIRS 


105 


that  the  base  must  absorb  so  large  a  proportion  of  the  thrust  that 
very  little  is  transmitted  to  the  sides  of  the  canyon.  The  exact 
determination  of  the  proportion  transmitted  in  the  higher  rings 
is  an  indeterminate  problem,  and  the  only  safe  method  is  to  assume 
with  regard  to  tangential  arch  stress  that  the  arch  stands  clear  of 


RAD.  150' 


Fig.  60.    Section  of  Pathfinder  Dam 

the  base.  This  will  leave  a  large  but  indeterminate  factor  of  safety 
and  enable  the  adoption  of  a  high  value  for  s,  the  maximum  unit 
stress. 

The  profile  of  the  dam  is  nearly  equiangular  in  outline.  This 
is  necessary  in  so  high  a  dam  in  order  to  bring  the  vertical  resultants 
(W)  R.  E.  and  (N)  R.  F.  as  near  the  center  as  possible  with  the 
object  of  bringing  the  ratio  of  maximum  to  mean  stress  as  low  as 
possible. 

The  estimation  of  the  exact  positions  of  W  and  of  N  is  made 
analytically  as  below. 


106  DAMS  AND  WEIRS 

There  are  only  two  areas  to  be  considered,  that  of  the  water 
overlying  the  inclined  back  (v)  and  that  of  the  dam  itself  (W). 
Dividing  v  by  2J  (the  assumed  specific  gravity  of  the  material), 
reduces  it  to  an  equivalent  area  of  concrete  or  masonry. 


104. 

W=  —  X  210  =  10920  =  768  tons 

Total,  or  N=  12390  =  871  tons 

Using  formula  (7),  Part  I,  the  e.g.  of  W  is  50.8  distant  from  the 
toe  of  the  profile,  then  q  or  the  distance  of  the  incidence  of  W  from 

the  center  point  of  the  base  is  50.8  —  —  =  3.8. 

2 

W        768 

The  value  of  $1,  or  the  mean  unit  stress  is  —  ,  or  —  —  =8.1  tons 

o          94 

.  6<7         .6X3.8  mW 

and    w  =  H—  r  =  H  —     -  =  1.24;    then    s=——  =  1.24X8.1  =10.1 
b  94  b 

tons. 

For  Reservoir  Full,  to  find  the  position  of  N,  moments  will  be 
taken  about  the  toe  as  follows 

Moment  of    v  =  103X83.5=  8600 

Moment  of  W  =  768  X  50.8  =  39014 

Total  #  =  871  =47614 

47614  94     „.,  #871 

then  x—  =54.6;  whence  q  =  54.6  —  -  =  7.6  feet  and  SI  =  -T  =  — 

871  •  2t  o       y4 

=  9.26.  By  formula  (9),  Part  I,  m  =  l+ 

C/T: 

5  =  9.26X1.48  =  13.7  tons 

From  this  it  is  evident  that  the  unit  stress  in  the  base,  due  to 
vertical  load  only,  is  a  high  figure.  It  could  be  reduced  by  still 
further  inclining  the  back;  on  the  contrary,  if  the  back  were  vertical 
N  would  equal  W.  Let  this  latter  case  be  considered.  The  distance 
of  the  c.  g.  of  the  profile  from  the  heel  will  then  be  by  formula 
(7a),  Part  I 

-81-68 


94 
and  the  value  of  q  will  be  —  —  31.66  =  15.33  feet 


DAMS  AND  WEIRS 


107 


Si  as  before  =  —  =  8.1  tons 
b 


Then 


09 

^- 
94 


and  5  =  8.1X1.98  =  16  tons 


This  stress  is  greater  than  that  of  N  in  the  previous  working  which 
proves  that  the  forward  tilt  given  to  these  high  dams  is  necessary 
to  reduce  the  maximum  unit  stress  on  the  base  to  a  reasonable  limit. 
A  more  equiangular  profile  would  give  even  better  results. 

85.  Shoshone  Dam.  The  Shoshone  dam,  Fig.  61,  is  designed 
on  lines  identical  with  the  last  example.  It  has  the  distinction  of 
being  the  highest  dam  in  the  world  but  has  recently  lost  this 


-108'- 
Fig.  61.     Profile  and  Force  Diagram  for  Shoshone  Dam 

preeminence,  as  the  Arrow  Rock,  quite  lately  constructed,  Fig.  37, 
Part  I,  is  actually  35  feet  higher.  This  work  is  also  in  the  United 
States.  The  incidents  of  the  resultants  Reservoir  Empty  and 
Reservoir  Full,  which  will  be  explained  later,  have  been  shown 
graphically,  and  the  analytical  computation  is  given  below.  The 
vertical  forces  taken  from  left  to  right  are  (1),  area  6480;  (2), 
14,450;  (3),  water  overlying  back,  reduced  area  1880;  total  22,810. 

Taking  moments  about  the  toe  of  the  base,  the  distance  of 
(1)  is  54  feet,  of  (2)  calculated  by  formula  (7),  Part  I,  is  58.3,  and 
of  (3)  is  95  feet,  roughly. 


108  DAMS  AND  WEIRS 

Then   (6480  X  54)  +  (14450  X  58.3)  +  (1880  X  95)  =22810Xz. 
.'.     #  =  60  feet,  nearly. 

1 0S 
The  value  of  q  for  N  then  is  60 —  =6  feet. 


Now  Sl  =  ^-  =  ~    -  =211,  and  by  formula  (9),  Part  I,  s=  211 X 
b        108 

(108+36)  281X2.4    01 

-=281  square  feet  = — —  =  21  tons,  nearly. 

1 08  32 

The  maximum  arch  unit  stress  by  formula  (22)  is  as  follows :  the 
radius  of  the  extrados  of  the  base  being  197  feet  the  fraction—  =—— 

2HW  2X245X1 

=  .55  and  H  =  245  therefore  s  =  — 


.55X1.45X32 


|g=  19.2  tons. 

Below  the  level  60  ft.  above  base,  the  stress  on  the  arch  does 
not  increase.  The  arch  stress  is  less  than  that  due  to  vertical 
pressure  N.  This  base  should  undoubtedly  have  been  widened, 
the  battered  faces  being  carried  down  to  the  base,  not  cut  off  by 
vertical  lines  at  the  60-foot  level. 

Center  of  Pressure — New  Graphical  Method.  In  order  to  find  the 
center  of  pressure  in  a  case  like  Fig.  61,  where  the  lines  of  forces  (1) 
and  (2)  are  close  together,  the  ordinary  method  of  using  a  force  and 
funicular  polygon  involves  crowding  of  the  lines  so  that  accuracy  is 
difficult  to  attain.  Another  method  how  will  be  explained  which  is 
on  the  same  principle  as  that  of  the  intersection  of  cross  lines  used 
for  finding  the  c.  g.  of  a  trapezoid. 

In  Fig.  61,  first  the  c.g.'s  of  the  three  forces  are  found  (1)  the 
water  pressure  area  divided  by  p  or  2.4  which  equals  1880  square 
feet,  (2)  the  upper  trapezoidal  part  of  the  dam  area  14,450,  and  (3) 
the  lower  rectangular  area  6480.  Then  (1)  is  joined  to  (2)  and  this 
line  projected  on  one  side  in  any  location  as  at  b  in  Fig.  61a. 

From  a,  ac  is  set  off  horizontally  equal  to  (2)  or  14,450  and 
from  b,  bd  is  drawn  equal  to  (1)  or  1880;  cd  is  then  drawn  and  its 
intersection  with  ab  at  e  gives  the  position  of  the  resultant  1-2,  which 
can  now  be  projected  on  the  profile  at  G.  To  obtain  the  resultant 
of  the  components  (1-2)  with  (3)  the  line  (7-3  is  drawn  on  the  profile 


DAMS  AND  WEIRS 


109 


and  a  parallel  to  it  drawn  from  e  on  Fig.  6 la,  intersecting  the  hori- 
zontal through  (3)  at  /.  From  e,  eg  is  laid  off  horizontally  equal  to 
(3)  or  6480  and  from/,  fh  equal  to  (1+2)  or  1880+14,450  =  16,330. 
hg  is  then  drawn  and  its  intersection  with  ef  at  j  is  the  centroid  of  the 
three  forces,  which  projected  on  the  profile  to  Gi  on  the  line  (7-3  gives 
the  location  of  the  vertical  resultant  of  1+2+3. 

86.    Sweetwater  Dam.    The  profile  of  the  Sweetwater  dam 
in  California  is  given  in  Fig.  62.    The  original  crest  of  the  dam 


Fig.  62.     Graphical  Analysis  of  Sweetwater 
Dam,  California 

was  at  El.  220,  or  95  feet  above  the  base.  Under  these  conditions 
the  dam  depended  for  its  stability  on  its  arched  plan.  If  con- 
sidered as  a  gravity  dam  with  allowable  tension  at  the  heel,  the 
vertical  pressure  area  is  the  triangle  ofo,  here  #  =  16.5  and  m  works 

out  to   3.15.     N  =  226  tons   and   6  =  46   feet   whence   s  =  ^  = 


3.15X226 
46 


=  15.5  tons  which  is  set  down  from  a  to  b. 


2N 

The  tension  at  the  heel  =  s2 r-  =  15.5  —  9.82  =  5.7  tons 

0 


110 


DAMS  AND  WEIRS 


E 


DAMS  AND  WEIRS 


111 


Value  of  S  When  Heel  Is  Unable  to  Take  Tention.  If  the  heel  is 
unable  to  take  tension,  the  pressure  triangle  will  then  be  adc  in  which 
ac  =  3  times  the  distance  of  the  incidence  of  R  from  the  toe,  or 
3X6.5  =  19.5  feet  and  s  is  ob- 
tained by  the  following  formula 
4      N  or  W 


here  s  = 


\        b-2q 
4X226 


(24) 


=  23.2  tons 


3X46-33 

This  dam  has  lately  been  raised 
to  El  240,  or  by  20,  feet  and  by 
the  addition  of  a  mass  of  concrete 
at  the  rear  transformed  into  a 
gravity  dam.  The  resultant  due 
to  this  addition  is  RI  on  the 
diagram,  s  works  out  to  10.6 
tons  and  there  is  no  tension  at 
the  heel.  Any  bond  between  the 
new  wall  and  the  old  has  been 
studiously  avoided.  The  new 
work  is  reinforced  with  cross  bars  and  the  rear  mass  tied  into  the 
superstructure.  Fig.  63  is  a  plan  of  the  dam  as  altered. 


Fig.  64.     Profile  of  Barossa  Dam 


Fig.  65.     Site  Plan  of  Barossa  Dam 


87.  Barossa  Dam.  This  dam,  Fig.  64,  is  an  Australian  work, 
and  although  of  quite  moderate  dimensions  is  a  model  of  good 
and  bold  design. 


112 


DAMS  AND  WEIRS 


RAD.  100' 


The  back  is  vertical  and  the  fore  batter  is  nearly  1  in  2.7.  The 
outline  is  not  trapezoidal  but  pentagonal,  viz,  a  square  crest  imposed 
on  a  triangle,  the  face  joined  with  the  hypothenuse  of  the  latter  by 
a  curve.  The  crest  is  slender,  being  only  4J  feet  wide,  but  is  strength- 
ened by  rows  of  40-pound  iron  rails,  fished  together,  built  into  the 
concrete.  The  maximum  arch  stress  works  out  to  17J  tons,  the 
corresponding  vertical  stress  on  base  to  6|  tons.  Fig.  65  is  a  site 
plan  of  the  work. 

88.  Lithgow  Dam.  Another  example  very  similar  to  the  last 
is  the  Lithgow  dam,  No.  2,  Fig.  66.  The  arch  stress  in  this  works 
out  by  the  short  formula  to  nearly  13  tons;  the  radius  is  only  100 
feet,  the  vertical  stress  works  out  to  7  tons. 
Arched  dams  abut  either  on  the  solid 
rocky  banks  of  a  canyon  or  else  on  the  end  of 
a  gravity  dam.  In  cases  where  a  narrow  deep 
central  channel  occurs  in  a  river,  this  por- 
tion can  advantageously  be  closed  by  an 
arched  dam,  while  the  flanks  on  which  the 
arch  abuts  can  be  gravity  dams  aligned 
tangential  to  the  arch  at  each  end.  The  dam 
will  thus  consist  of  a  central  arch  with  two 
inclined  straight  continuations.  The  plan  of 
the  Roosevelt  dam,  Fig.  26,  Part  I,  will  give 
an  idea  of  this  class  of  work. 

89.     Burrin  Juick  Subsidiary  Dam.     In 

Fig.  67  is  shown  the  profile  of  a  temporary  reinforced  arched  dam 
for  domestic  water  supply  at  Barren  Jack,  or  Burrin  Juick,  Australia. 
The  reinforcement  consists  of  iron  rails.  The  unit  arch  pressure 
at  the  base  works  out  to  21  tons,  nearly.  Reinforcement  of  perma- 
nent dams  down  to  the  base  is  not  desirable,  as  the  metal  may  cor- 
rode in  time  and  cause  failure,  although  the  possibility  is  often 
stoutly  denied.  The  main  Burrin  Juick  dam  is  given  in  Part  I, 
Fig.  36. 

90.  Dams  with  Variable  Radii.  The  use  of  dams  of  the  type 
just  described,  is  generally  confined,  as  previously  noted,  to  narrow 
gorges  with  steep  sloping  sides  in  which  the  length  of  the  dam  at  the 
level  of  the  bed  of  the  canyon  is  but  a  small  proportion  of  that  at  the 
crest.  The  radius  of  curvature  is  usually  fixed  with  regard  to  the 


Fig.  66. 


Profile  of  Lithgow 
Dam 


DAMS  AND  WEIRS 


113 


length  of  chord  at  the  latter  level,  consequently  at  the  deepest 

level,  the  curvature  will  be  so  slight  that  arch  action  will  be  absent 

and  the  lower  part  of  the  dam  will  be  subject  to  beam  stresses,  i.e., 

to  tension  as  well  as  compression.     In  order  to  obviate  this,  in 

some  recent  examples  the  radius  of  curvature  at  the  base  is  made 

less  than  that  at  the  crest,  and  all  the  way  up,  the  angle  subtending 

the  chord  of  the  arc,  which  is  variable 

in    length    retains    the    same    measure 

throughout.     This  involves  a  change  in 

the  radius  corresponding  to  the  variable 

span  of  the  arch.    The  further  advantage 

is  obtained,  of  reduction  in  the  unit  stress 

in  the  arch  ring  and  in  rendering  the 

stress    more    uniform    throughout.     In 

very  high  dams,  however,  the  base  width 

cannot  be  much  reduced  as  otherwise  the 

limit  stress  due  to  the  vertical  loading 

will  be  exceeded.    This  arrangement  of 

varying  radii  is  somewhat  similar  to  that 

used  in  the   differential    multiple    arch 

given  later. 

MULTIPLE    ARCH    OR 

HOLLOW    ARCH 

BUTTRESS  DAMS 

91.    Multiple  Arch  Generally  More 
Useful  Than  Single  Arch  Dams.    It  is 

evident  that  a  dam  which  consists  of  a 
single  vertical  arch  is  suitable  only  for  a 
narrow  gorge  with  rock  sides  on  which  the 
arch  can  abut,  as  well  as  a  rock  bed;  consequently  its  use  is  strictly 
limited  to  sites  where  such  conditions  are  obtainable.  A  rock 
foundation  is  also  essential  for  gravity  dams,  the  unit  compression 
on  the  base  of  which  is  too  high  for  any  material  other  than  rock. 
The  advantages  inherent  in  the  vertical  arch,  which  are  con- 
siderable, can  however  be  retained  by  use  of  the  so-termed  multiple 
or  scallop  arched  dam.  This  consists  of  a  series  of  vertical  or 
inclined  arches,  semicircular  or  segmental  on  plan,  the  thrust  of 


Fig.  67.     Profile  of  Burrin  Juick 
Subsidiary  Dam 


114  DAMS  AND  WEIRS 

which  is  carried  by  buttresses.     The  arrangement  is,  in  fact,  iden- 
tical with  that  of  a  masonry  arched  bridge.     If  the  latter  be  con- 
sidered as  turned  over  on  its  side,  the  piers  will  represent  the  but-  ; 
tresses.     In  the  case  of  a  wide  river  crossing,  with  a  bed  of  clay,    , 
boulders,  or  sand,  the  hollow  buttressed  and  slab  buttressed  dams 
are  the  only  ones  that  can  well  be  employed  with  safety.     The  wide 
spread  that  can  be  given  to  the  base  of  the  structure  in  these  two 
types  enables  the  unit  pressure  on  the  base  to  be  brought  as  low  as 
from  2  to  4  tons  per  square  foot. 

As  has  already  been  noticed  in  section  78,  the  arch  is  peculiarly 
well  suited  for  economical  construction.  This  is  due  to  the  fact 
that  the  liquid  pressure  to  which  the  arch  is  subjected  is  normal 
to  tjie  surface  and  radial  in  direction.  The  pressure  lines  in  the 
interior  of  the  arch  ring  correspond  with  its  curvature  and  con- 
sequently the  arch  can  only  be  in  compression;  thus  steel  reinforce- 
ment is  unnecessary  except  in  a  small  degree  near  the  crest  in  order 
to  care  for  temperature  stresses.  In  slab  dams,  on  the  other  hand, 
the  deck  is  composed  of  flat  slabs  which  have  to.  be  heavily  rein- 
forced. The  spacing  of  the  buttresses  for  slabs  is  limited  to  15 
to  20  feet,  whereas  in  hollow  arch  dams  there  is  practically  no 
limit  to  the  spans  which  may  be  adopted.  Another  point  is,  that 
the  extreme  compressive  fiber  stress  on  the  concrete  in  deck  slabs 
is  limited  to  five  hundred  to  six  hundred  and  fifty  pounds  per  square 
inch;  in  an  arch,  on  the  other  hand,  the  whole  section  is  in  com- 
pression which  is  thereby  spread  over  a  much  greater  area.  For  the 
reasons  above  given  the  arch  type  now  under  consideration  should 
be  a  cheaper  and  more  scientific  construction  than  the  slab  type  in 
spite  of  the  higher  cost  of  forms. 

92.  Mir  Alam  Dam.  The  first  example  given  is  that  of  the 
Mir  Alam  tank  dam,  Fig.  68.  This  remarkable  pioneer  structure  was 
built  about  the  year  1806,  by  a  French  engineer  in  the  service  of 
H.  H.  the  Nizam  of  Hyderabad  in  Southern  India.  The  alignment 
of  the  dam  is  on  a  wide  curve  and  it  consists  of  a  series  of  vertical 
semicircular  arches  of  various  spans  which  abut  on  short  buttress 
piers,  Fig.  69.  The  spans  vary  from  83  to  138  feet,  the  one  in  Fig. 
68  being  of  122  feet.  The  maximum  height  is  33  feet.  Water 
has  been  known  to  overtop  the  crest.  The  length  of  the  dam  is 
over  3000  feet. 


DAMS  AND  WEIRS 


115 


On  account  of  the  inequality  of  the  spans,  the  adoption  of  the 
semicircular  form  of  arch  is  evidently  a  most  judicious  measure, 
for  the  reason  that  an  arch  of  this  form  under  liquid  pressure  exerts 
o  lateral  thrust  at  the  springing.  The  water  pressure  being  radial 
in  direction,  cross  pressure  in  the  half  arches  in  the  line  of  the  spring- 
ing is  balanced  and  in  equilibrium.  Whatever  thrust  is  exerted 
is  not  in  the  direction  of  the  axis  of  the  dam  but  that  of  the  buttress 
piers.  On  the  other  hand,  if  the  arches  were  segmental  in  outline 
the  terminal  thrust  is  intermediate  between  the  two  axes,  and  wheu 
resolved  in  two  directions  one  component  acts  along  the  axis  of 


Fig.  68.     Plan  of  One  Arch  of  Mir  Alam  Dam 
This  remarkable  pioneer  dam  was  built  in  1806,  and  consisted  of  21  such  arches. 

the  dam.  This  has  to  be  met,  either  by  the  abutment,  if  it  is  an 
end  span,  or  else  by  the  corresponding  thrust  of  the  adjoining  half 
arch.  The  other  component  is  carried  by  the  buttress;  therefore, 
if  segmental  arches  are  used,  the  spans  should  be  equal  in  order  to 
avoid  inequality  of  thrust.  Longer  buttresses  will  also  be  requisite. 
The  whole  of  this  work  is  built  of  coursed  rubble  masonry  in  lime 
mortar;  the  unit  stress  in  the  arch  ring  at  the  base,  using  the  short 

f         ,    ,-IN  (RHw)  68X33X1 

1  ormula  (21),  -  —  —  -  works  out  to  —  -  —  —  —  =  5  tons,  nearly.     The 


dam?  therefore,  forms  an  economical  design. 


116 


DAMS  AND  WEIRS 


The  buttress  piers  are  shown  in  section  in  Fig.  70,  the  section 
being  taken  through  AB  of  Fig.  68.  In  this  work  the  buttress 
piers  are  very  short,  projecting  only  25  feet  beyond  the  spring  line 
of  the  arches,  and  being  altogether  only  35  feet  long.  This  length 
and  the  corresponding  weight  would  clearly  be  inadequate  to  with- 


Fig.  69.     Plan  of  Entire  Mir  Alam  Dam 


stand    the    immense    horizontal    thrust    which    is    equivalent    to 
IP,    =332X1X146 
2  W 


=  2500  tons,  nearly. 


2X32 

It  is  evident  that  if  the  buttress  pier  slides  or  overturns,  the 
arches  behind  it  must  follow,  for  which  reason  the  two  half  arches  and 
the  buttress  pier  cannot  be  considered  as  separate  entities  but  as 
actually  forming  one  whole,  and  consequently  the  effective  length 
of  the  base  must  extend  from  the  toe  of  the  buttress  right  back 
to  the  extrados  of  the  two  adjoining  arches.  At,  or  a  little  in  the 


Section  f\-B 

^^'-oH- 


Present  Surfai 


Fig.  70.     Section  of  Buttress  Pier  of  Mir  Alam  Dam  Taken  through  AB  of  Fig.  68 

rear  of  the  spring  line,  the  base  is  split  up  into  two  forked  curved 
continuations.  The  weight  of  these  arms,  i.e.,  of  the  adjoining 
half  arches,  has  consequently  to  be  included  with  that  of  the  but- 
tress proper  when  the  stability  of  the  structure  against  overturning 
or  sliding  is  estimated. 


DAMS  AND  WEIRS 


117 


93.  Stresses  in  Buttress.  In  the  transverse  section,  Fig.  71, 
taken  through  CD  of  Fig.  68,  the  graphical  calculations  establish 
the  fact  that  the  resultant  line  R  intersects  the  base,  thus  lengthened, 
at  a  point  short  of  its  center;  the  direction  of  the  resultant  R  is 
also  satisfactory  as  regards  the  angle  of  frictional  resistance. 

Ri  is  the  resultant  on  the  supposition  that  the  buttress  is 
nonexistent.  Its  incidence  on  the  base  proves  that  the  arch  is 
stable  without  the  buttress,  which  is  therefore  actually  superfluous. 
With  regard  to  sliding  on  the  base,  P  =  2500  and  W  =  6828  tons. 


Fig.  71.     Transverse  Section  of  Mir  Alam  Dam  Taken  through  CD,  Fig.  68 

The  coefficient  of  friction  being  .7  the  factor  of  safety  against  sliding 
is  nearly  2.  If  the  arch  were  altered  on  plan  from  a  semicircle 
to  a  segment  of  a  circle,  the  radius  wrould  of  necessity  be  increased, 
and  the  stress  with  it;  a  thicker  arch  would,  therefore,  be  required. 
This  would  not  quite  compensate  for  the  reduced  length  of  arch, 
but  on  the  other  hand,  owing  to  the  crown  being  depressed,  the 
effective  base  width  would  be  reduced  and  would  have  to  be  made 
good  by  lengthening  the  buttress  piers.  What  particular  dis- 
position of  arch  and  buttress  would  be  the  most  economical  is  a 


118 


DAMS  AND  WEIRS 


matter  which  could  only  be  worked  out  by  means  of  a  number  of 
trial  designs.  The  ratio  of  versed  sine  to  span  should  vary  from 
J  to  ^.  Arcs  subtending  from  135  to  120  degrees  are  stated  to  be 
the  most  economical  in  material. 

94.     Belubula  Dam.    There  are  not  as  yet  very  many  modern 
examples  of  arch  buttress  dams,  but  each  year  increases  their  num- 


flrch 


\ 


Fig.  72.     Profile  Sections  and  Force  Diagram  for  Belubula  Dam,  New  South  Wales 

ber.  The  Mir  Alam  dam  has  remained  resting  on  its  laurels  without 
a  rival  for  over  100  years,  but  the  time  has  come  when  this  type 
is  being  largely  adopted.  Fig.  72  shows  an  early  example  of  a 
segmental  panel  arch  dam.  It  is  the  Belubula  dam  in  New  South 
Wales.  The  arch  crest  is  37  feet  above  the  base,  very  nearly  the 
same  as  in  the  last  example.  The  arches,  which  are  inclined  60  degrees 
to  the  horizontal  are  built  on  a  high  solid  platform  which  obliterates 


DAMS  AND  WEIRS  119 

inequalities  in  the  rock  foundation.  This  platform  is  16  to  23  feet 
high,  so  that  the  total  height  of  the  dam  is  over  50  feet.  The  spans 
are  16  feet,  with  buttresses  12  feet  wide  at  the  spring  line,  tapering 
to  a  thickness  of  5  feet  at  the  toe ;  they  are  40  feet  long.  The  buttress 
piers,  which  form  quadrants  of  a  circle  in  elevation,  diminish  in 
thickness  by  steps  from  the  base  up,  these  insets  corresponding 
with  similar  ones  in  the  arch  itself.  These  steps  are  not  shown 
in  the  drawing;  the  arch  also  is  drawn  as  if  in  one  straight  batter. 
The  arches  are  elliptical  in  form,  and  the  spandrels  are  filled  up 
flush  with  the  crown,  presenting  a  flat  surface  toward  the  water. 

Some  of  the  features  of  this  design  are  open  to  objection :  First, 
the  filling  in  of  the  arch  spandrels  entirely  abrogates  the  advantage 
accruing  to  arches  under  liquid  pressure.  The  direction  of  the  water 
pressure  in  this  case  is  not  radial  but  normal  to  the  rear  slope,  thus 
exactly  reproducing  the  statical  condition  of  a  horizontal  arch 
bridge.  The  pressure,  therefore,  increases  from  the  crown  to  the 
haunches  and  is  parabolic,  not  circular,  in  curvature.  The  arches 
should  have  been  circular,  not  elliptical,  and  the  spandrels  left 
empty  to  allow  of  a  radial  pressure  which  partly  balances  itself. 
Second,  the  stepping  in  of  the  intrados  of  the  arch  complicates  the 
construction.  A  plain  batter  would  be  easier  to  build,  particularly 
in  concrete.  Third,  the  tapering  of  the  buttress  piers  toward  the 
toe  is  quite  indefensible;  the  stress  does  not  decrease  but  with  the 
center  of  pressure  at  the  center  of  the  base  as  in  this  case,  the 
stress  will  be  uniform  throughout. 

95.  Inclination  of  Arch  to  Vertical.  The  inclination  of  the  axis 
of  the  arch  to  the  vertical  is  generally  a  desirable,  in  fact,  a  necessary 
feature  when  segments!  arched  panels  are  used;  the  weight  of  water 
carried  is  of  value  in  depressing  the  final  resultant  line  to  a  suitable 
angle  for  resistance  to  shearing  stress.  As  noted  in  section  90, 
the  weight  of  the  water  overlying  the  arch  does  not  increase  the 
unit  stress  in  the  arch  ring.  Consequently,  any  inclination  of 
axis  can  be  adopted  without  in  any  way  increasing  the  unit  stresses 
due  to  the  water  pressure. 

When  an  arch  is  vertical  it  is  clear  that  the  water  pressure  is 
all  conveyed  to  the  abutments  and  the  weight  of  the  arch  to  its 
base.  When  an  arch  lies  horizontally  under  water  pressure  both 
the  weight  of  the  water  and  that  of  the  arch  itself  are  conveyed 


120  DAMS  AND  WEIRS 

to  the  abutment;  when  in  an  intermediate  position  part  of  the  weight 
of  the  arch  is  carried  to  the  base  and  part  to  the  abutments. 

With  regard  to  water  pressure,  the  thrust  being  normal  to 
the  extrados  of  the  arch  the  whole  is  carried  by  the  abutments. 
In  the  case  of  arches  which  do  not  overreach  their  base  the  weight 
of  water  overlying  the  inclined  back  is  conveyed  to  the  base.  In 

any  case  the  unit  stress  in  the  arch  -  — cannot  exceed  that  due 

to  horizontal  thrust.  The  total  water  pressure  is  greater  with  an 
inclined  back,  as  the  length  of  surface  acted  on  is  increased.  In 
the  diagram,  Fig.  72a,  the  vertical  load  line  W  represents  the 
weight  of  one  unit  or  one  cubic  foot  of  the  arch  ring  which  is  equal 
to  w p.  This  force  is  resolved  in  two  directions,  one  p,  parallel 
to  the  axis  of  the  arch,  and  the  other  n,  normal  to  the  former.  The 
force  n  =  W  sin  6,  6  being  the  inclination  of  the  arch  axis  to  the 
vertical  and  p  =  W  cos  6.  The  unit  stress  developed  by  the  radial 
force  n  is  similar  to  that  produced  by  the  water  pressure  which  is 
also  radial  in  direction  and  is  R in;  but  RI,  the  radius  in  this  case, 
is  the  mean  radius,  the  pressure  being  internal,  not  external.  The 
unit  stress  Si  will  then  be 

Si  =  RiWp  sin  6  (25) 

When  6  is  30°  sin  0  =  4^  when  45°'  sin  6==^' 
2  o 

It  will  easily  be  understood  that  this  unit  stress  due  to  n  does 
not  accumulate,  but  is  the  same  at  the  first  foot  depth  of  the  arch 
as  it  is  at  the  bottom;  the  width  of  the  lamina  also  does  not  affect 
it.  However,  the  component  p  does  accumulate,  and  the  expression 
wp  cos  6  should  be  multiplied  by  the  inclined  height  HI,  lying 
above  the  base  under  consideration.  As  H}  =  H  sec  d,  the  unit 

compressive  stress  at  the  base  will  be — - ,  in  which  b\  is  the  mean 

width  of  the  arch.  If  the  arch  were  a  rectangle,  not  a  trapezoid, 
s  would  equal  Hwp  simply. 

96.  Ogden  Dam.  The  Ogden  dam,  the  profile  and  sectional 
details  of  which  are  shown  in  Fig.  73,  is  a  notable  example  of  the 
arch  and  buttress  type.  Its  height  is  100  feet.  The  inclination 
of  the  arches  is  less  than  \  to  1,  or  about  25  degrees  to  the  vertical. 


DAMS  AND  WEIRS 


121 


The  profile  of  the  buttress  is  equinangular  except  for  a  small  out- 
throw  of  the  toe.  On  the  whole  it  must  be  pronounced  a  good 
design,  but  could  be  improved  in  several  particulars.  For  example, 
the  arch  is  unnecessarily  thick  at  the  crest,  and  could  well  be  reduced 
from  6  to  2  feet,  thus  effecting  considerable  economy.  The  designers 
were  evidently  afraid  of  the  concrete  in  the  arch  leaking,  and  so 
overlaid  the  extrados  with  steel  plates.  A  greater  thickness  of 
arch  causing  it  to  possess  less  liability  to  percolation  under  pressure, 
could  have  been  provided  by  increasing  the  span  and  radius  of  the 


/ 

"N     /^\     S* 

s  ^ 

-3JO\ 

? 

•16* 

^pt* 

1 

w  - 

y 

T 

•  — 

I   U 

OM  AA   ^  SECTION  OttBB     PL  Att  OVERALL 

Fig.  1Z.     Profile  and  Sections  of  Ogden  Dam 


arches.  The  design  consequently  would  be  improved  by  adopting 
larger  spans,  say  100  feet;  buttresses,  say,  25  feet  thick,  their  length 
being  dependent  on  the  width  of  base  required  to  provide  sufficient 
moment  of  resistance;  and  further,  the  inclination  of  the  arches 
might  require  increasing  to  bring  the  center  of  pressure  at,  or  close 
to  the  center  of  the  buttress.  The  finish  of  the  crest  by  another 
arch  forming  a  roadway  is  an  excellent  arrangement,  and  is  well 
suited  for  a  bulkhead  dam;  for  an  overfall,  on  the  other  hand,  the 
curved  crest  is  preferable  on  account  of  the  increased  length  of 
overflow  provided.  The  stress  diagram  shows  that  the  value  of  the 


122  DAMS  AND  WEIRS 

vertical  load  N  is  155,000  cubic  feet  or  10,598  tons,  p  being  taken 
at  2J.  The  incidence  of  R  on  the  base,  is  5  feet  from  the  center, 
whence  #  =  5,  and  by  formula  (9),  Part  I 

N^f*  .  6A     10598  x/14° 

s= AX(I+  b)= no->o6xTTo=8-91  tons 

the  dimensions  of  A,  the  area  of  the  base,  being  110X16  feet.  The 
pressure  on  the  arch  ring  at  the  base  by  the  short  formula  works 

24X100    n, 

out  to  -——  =  9.4  tons. 
0X0^ 

TU                      t^     j               t                                      104,500 
I  he  contents  ot  the  dam  per  foot  run  amounts  to   — = 

48 

2,177  cubic  feet;  that  of  a  gravity  dam  would  be  about  3,500  cubic 
feet  per  foot  run,  making  a  saving  in  favor  of  the  arched  type  of 
nearly  30  per  cent.  With  a  better  disposition  of  the  parts  as  indi- 
cated above,  the  saving  would  be  increased  to  40  or  50  per  cent. 
Actually  the  saving  amounted  to  only  12  per  cent;  this  was  owing 
to  the  steel  covering  which,  as  we  have  seen,  could  have  been 
dispensed  with. 

97.  Design  for  Multiple  Arch  Dam.  Fig.  74  is  a  design  for 
a  segmental  arch  panel  dam,  or  rather,  weir.  The  height  of  the 
crest  is  64  feet  above  base  with  5  feet  of  water  passing  over;  the 
apex  of  the  triangle  of  water  pressure  will  then  be  69  feet  above 
the  base.  The  inclination  given  the  axis,  which  is  coincident  with 
that  of  the  spring  line  and  the  intrados,  is  60  degrees  with  the  horizon. 

In  designing  such  a  work,  the  following  salient  points  first 
require  consideration. 

(1)  Width  of  Span.    This,  it  is  deemed  for  economical  reasons 
should  be  not  less  than  the  height  of  crest  unless  the  state  of  the 
foundation  requires  a  low  unit   stress.    In  the  Mir   Alam  dam 
the  span  is  over  four  times  the  depth  of  water  upheld.     In  the  present 
case  it  will  be  made  the  same,  that  is,  64  feet. 

(2)  Thickness  of  Buttress  Piers.    As  with  bridge  piers,  the 
width  should  be  at  least  sufficient  to  accommodate  the  skew-backs 
of  the  two  arches;  a  width  of  12  feet  or  about  i  span  will  effect  this. 

(3)  Radius  and   Versed  Sine.    The  radius  will  be  made  40 
feet;  this  allows  a  versed  sine  of  J  span,  or  16  feet,  which  is  con- 
sidered to  be  about  the  flattest  proportion  to  afford  a  good  curva- 


UVdS  JO  'T3 


124  DAMS  AND  WEIRS 

ture,  the  greater  the  length  of  the  arc,  the  more  its  condition  will 
approximate  to  that  of  a  circular  arch,  under  liquid  pressure. 

(4)  Thickness  of  Arch.  This  must  first  be  assumed,  as  its 
thickness  depend*  on  R,  the  radius  of  the  extrados,  as  well  as  on 
the  value  assigned  to  si,  the  limiting  pressure.  This  latter  will 
be  fixed  at  below  15  tons,  a  value  by  no  means  excessive  for  arches 
under  liquid  pressure.  With  a  base  width  of  7  feet,  the  radius 
of  the  extrados  will  be  47  feet.  The  base  will  be  considered,  not 
at  the  extreme  depth  of  64  feet  below  crest,  but  at  the  point  marked 
D,  where  a  line  normal  to  the  base  of  the  inclined  intrados  cuts 
the  extrados  of  the  arch.  //  will,  therefore,  be  60  feet,  allowing 
for  the  reverse  pressure.  The  stress  due  to  the  water  pressure, 
using  the  short  formula  (21),  section  78,  will  be 

RHw     47X60X1 


To  this  must  be  added  that  due  to  the  weight  of  the  arch  ring  from 
formula  (25),  s^  =  Riicp  sin  6  (the  angle  6  being  30°  and  its  sine  =  J), 

43  5x3 
which  in  figures  will  be     ''          =1.6  tons,  the  total  stress  being 

a  trifle  over  14  tons.  The  7-foot  base  width  will  then  be  adopted. 
p  is  taken  as  2.4  and  wp  =  ^-0-  ton.  The  depth  of  water  producing 
this  pressure  is  taken  as  60,  not  as  65,  feet  which  is  (H+d),  the  reason 
being  that  the  reverse  pressure  due  to  the  tail  water,  which  must 
be  .at  least  level  with  the  water  cushion  bar  wall,  will  reduce  the 
effective  depth  to  60  feet,  during  flood  conditions, 

98.  Reverse  Water  Pressure.  The  influence  of  the  reverse 
pressure  of  water  is  much  more  considerable  when  hydrostatic 
pressure  alone  is  exerted  than  is  the  case  with  overturning  moment. 
In  the  case  of  an  upright  arch  acting  as  an  overfall  weir  the  pres- 
sure of  the  tail  water  effects  a  reduction  of  the  pressure  to  the  extent 
of  its  area.  Thus  if  A  be  the  area  of  the  upstream  water  pressure, 
and  a  that  of  the  downstream,  or  tail  water,  the  unbalanced  pressure 
will  be  their  difference,  or  A  —  a,  and  will  vary  as  the  square  of  their 
respective  depths.  When  overturning  moment  is  concerned,  the 
areas  have  to  be  multiplied  by  a  third  of  their  depths  to  represent 
the  moment  on  the  base.  The  difference  of  the  two  will  be  in 
that  case  as  the  cubes  of  their  respective  depths. 


DAMS  AND  WEIRS  125 

99.  Crest  Width  of  Arch.  The  crest  width  of  the  arch,  accord- 
ing to  formula  (23),  should  be  jV//  =  3f  feet,  nearly.  It  will  be 
made  3  feet,  with  a  stiffening  rib  or  rim  of  3  feet  in  width.  The 
crest  width  could  be  made  proportional  to  the  base  width,  say  .36, 
and  if  this  falls  below  2  feet,  reinforcement  will  be  required. 

The  length  of  the  pier  base  is  measured  from  the.  extrados  of 
the  arch,  the  two  half  arches  forming,  as  already  explained  in  section 
92,  a  forked  continuation  of  the  buttress  pier  base. 

The  battering  of  the  sides  of  the  pier  would  clearly  be  a  correct 
procedure,  as  the  pressure  diminishes  from  the  base  upward.  A 
combined  batter  of  1  in  10  is  adopted,  which  leaves  a  crest  width 
of  5.6  feet.  The  length  of  the  pier  base,  as  also  its  outline,  were 
determined  by  trial  graphical  processes,  with  the  object  of  maneu- 
vering the  center  of  pressure  as  near  that  of  the  base  as  pos- 
sible, so  as  to  equalize  the  maximum  and  the  mean  unit  stress  as 
much  as  possible.  This  has  been  effected,  as  shown  by  the  inci- 
dence of  the  final  resultant  on  the  elevation  of  the  buttress  pier. 

ICO.  Pressure  on  Foundations.  The  total  imposed  weight 
is  measured  by  N  in  the  force  diagram,  and  is  equivalent  to  150,000 
cubic  feet  of  masonry,  which  at  a  specific  gravity  of  2.4  is  equal  to 

=  11,250  tons.     The  average  pressure  is  this  quantity 

divided  by  the  area  of  the  base,  or  by  125X12  =  1500  square  feet, 
the  quotient  being  7J  tons,  nearly.  The  maximum  pressure  will 
be  the  same  owing  to  the  incidence  of  R  at  the  center  of  the  base. 
This  7|  tons  is  a  very  moderate  pressure  for  a  hard  foundation; 
if  excessive,  additional  spread  should  be  provided  or  else  the  spans 
reduced.  It  will  be  noticed  that  N  greatly  exceeds  W.  This  is 
due  to  the  added  weight  of  water  represented  by  the  inclination 
given  to  the  force  line  P,  which  represents  the  water  pressure. 

Economy  of  Multiple   Arches.    The   cubic  contents  per  foot 

A4  000 
run  work  out  to  — ^ —  =  850  cubic  feet,  nearly,  the  denominator  in 

the  fraction  being  the  distance  apart  of  the  centers  of  the  buttress  piers. 

2 
The  contents  of  a  gravity  weir  with  base  width  ^  (H+d)  and 

top  V//+d,  works  out  to  1,728  cubic  feet;  the  saving  in  material 
is  therefore  over  50  per  cent. 


126  DAMS  AND  WEIRS 

101.  Differential  Arches.  Fig.  75  is  a  study  of  a  differential 
buttress  arch  weir.  The  principle  of  the  differential  arch  consists 
in  the  radius  increasing  with  the  height  of  the  arch,  the  unit 
stress  is  thus  kept  more  uniform,  and  the  stress  area  corre- 
sponds more  closely  with  the  trapezoidal  profile  that  has  necessarily 
to  be  adopted,  than  is  the  case  when  a  uniform  radius  is  adopted  as 
in  Fig.  74. 

The  arches  are  supposed  to  stand  on  a  concrete  or  masonry 
platform  ten  feet  high  above  the  deepest  part  of  the  river  bed, 
so  that  sluices  if  required  could  be  provided  below  L.W.L.  which 
is  identical  with  the  floor  or  fore  apron  level.  The  height  is  35  feet 
to  crest  level.  The  depth  of  film  passing  over  the  crest  is  assumed 
at  5  feet  and  the  reciprocal  depth  of  tail  water  is  12  feet.  Graphical 
analyses  will  be  made  at  two  stages,  first,  when  water  is  at  crest 
level  and  the  river  channel  below  is  empty,  second,  at  full  flood. 
The  inclination  given  to  the  intrados  of  the  arch  is  3  vertical  to  2 
horizontal.  The  buttresses  are  placed  31  feet  centers,  allowing 
a  span  of  25  feet  at  base,  here  they  are  6  feet  wide,  tapering  to  2 
feet  at  crest.  The  span  of  the  arch  thus  gradually  widens  from 
25  to  29  feet.  The  versed  sine  of  the  arc  is  made  5  feet  at  base 
and  2\  feet  at  crest.  The  radii  at  these  positions  are  therefore  18.1 
and  43.3,  respectively,  measured  to  the  intrados  of  the  arch.  These 
radii  are  horizontal,  not  normal  to  the  intrados  as  in  Fig.  73,  and 
thus  vary  right  through  from  18.1  to  43.3  corresponding  to  the 
altered  versed  sine  which  decreases  from  5  to  2|  feet,  that  half 
way  up  being  22  feet. 

The  thickness  of  the  arch  at  base  is  made  2  feet. 

Arch   Unit  Stress.     Taking  the  base  radius  as  18.1,  the  base 

unit  stress  due  to  water  pressure  will  be  by  formula  (21),  s  =  — - — • 

o 

adding  that  due  to  the  transmitted  weight  of  the  arch,  formula  (25), 
s  =  Rwt  — +p  sin  6  I,    sin    6    being    .6,    the    expression    becomes 

IS.lw  I  Y+(2.4X.6)  1 ,  p  being  taken  at  2.4,  w  at  A  ton,  whence 

s  =  10.4  tons,  a  moderate  stress  for  a  vertical  arch. 

The  real  thickness  of  the  arch  is  more  like  2J  feet  than  2  feet 
as  properly  it  should  be  measured  horizontally,  not  normally. 


DAMS  AND  WEIRS 


127 


Upper  Ftirllo  be  Reinforced 
for  Temperafure  y Shock. 


Fig.  75.    Design  Diagrams  for  Differential  Buttress  Arch  Weir 


128  DAMS  AND  WEIRS 

Load  Line.     In  the  force  polygon  the  load  line  is  made  up  of 
five  weights:   (1)    that  of    the  overlying  water  has  a  content  of 

=  13860  cubic  feet,  equivalent  to  433  tons;   (2)  the  arch 


145  tons;  (3)  the  contents  of  the  pier  underlying  the  arch  is  found 
by  taking  the  contents  of  the  whole  as  if  the  sides  were  vertical 
and  deducting  the  pyramid  formed  by  the  side  batters.  Thus  the 

22x6X35 
contents  of  the  whole  is  -  -  -  =2310,  that  of  the  pyramid  is 

2i 

=513,  difference  1797,  or  135  tons;   (4)   the  weight  of 
o 

the  horizontal  arch  of  the  crest  of  the  weir,  12  tons;  (5)  the  contents 
of  the  buttress,  by  the  prismoidal  formula  comes  to  —  I  (Ai+  ^Am-\- 
Az)  in  which  AI  and  A2  are  areas  of  the  ends  and  Am  of  the  middle 
section.  Here  A  =  0,  ^m  =  yX  J-  =  52.5,  and  ^12  =  35X4  =  140; 

therefore  (5)  =  -Vx  35  X[0+  (4X52.5)  +  140]  =  2042  cubic  feet  equiv- 

alent to  153  tons. 

The  total  load  foots  up  to  878  tons. 

H2  1      C35^2 

P  the  horizontal  water  pressure  =  w—-X/  =  —  X-^-  X31  =593 

2  32        2 

tons. 

The  position  of  the  several  vertical  forces  is  obtained  as  follows  : 
That  of  1,  a  triangular  curved  prism  is  at  J  its  horizontal  width; 
of  2  is  found  by  formula  (7),  Part  I,  and  by  projection  of  this  level 
on  to  the  plan.  The  position  of  3  has  to  be  calculated  by  moments 
as  below. 

The  lever  arm  of  the  whole  mass  including  the  battered  sides 
is  at  J  width  from  the  vertical  end  of  7  J  feet  while  that  of  the  pyram- 
idal batter  is  at  J  the  same  distance,  or  5|  feet. 

The   statement   is   then 
2310X7J  =  (1797Xz)  +  (514x5.5)  whence  z  =  7.84  feet 

The  position  of  5,  the  battered  sloping  buttress  is  obtained  by 
taking  the  center  part  2  feet  wide  and  the  outer  side  batters  sep- 

1  35 

arately.    The  e.g.  of  the  former  is  at  —  the  length,  —  =  11§  from  the 

o  o 


DAMS  AND  WEIRS  129 

352 
vertical  end,  and  its  contents  are  -^-X2  =  1225  cubic  feet  =  92  tons. 

2i 

The  weight  of  the  whole  is  153  tons,  so  that  the  side  batters  will 
weigh  153-92  =  61  tons,  and  be  ^  =  8.75  feet  distant  from  the  end. 

Taking  moments  about  the  vertical  end,  we  have 
153z  =  (92  XI  1.67)  +  (61X8.75) 

1606     inr, 
x  =  —- —  =10.5  feet 
15o 

Therefore,  the  incidence  of  the  resultant  on  the  base  line  meas- 
ured 6.5  feet  upstream  from  the  center  point. 

N 
In  Fig.  75a,  N  =  87S  tons  and  —  =  14;  b  being  63  feet  and 

</  =  6.5  feet,  whence  ??i  =  1.62  and  the  stress  on  the  buttress,  if  only  1 
foot  wide,   =14X1.62  =  22.7  tons.     The  compression   at  the   toe 

9  N 

=— $  =  (28—22.7)  =5.3  tons.    These  quantites  have  now  to  be 

o 

divided  by  the  base  widths  to  obtain  the  unit  stresses,  which  are  as 

22.7  14  5.3 

follows:  at  heel,  -^-  =  3.8;  at  center,  —  =  2.3;  at  toe,  -^-=  2.6  'tons. 

This  stress  area  is  shown  hatched  in  Fig.  75f. 

This  stress  diagram  is  useful  as  showing  that  owing  to  the 
incidence  of  R  being  behind  the  center  point  the  total  stress  dimin- 
ishes toward  the  toe  of  the  buttress,  consequently  it  should  be 
tapered  on  plan,  as  has  been  done.  In  Fig.  74  it  has  been  shown 
that  the  stress  being  uniform  by  reason  of  the  incidence  of  R  at 
the  center  point  of  the  base,  the  buttress  has  been  made  rectangular 
in  plan  at  its  base.  The  indicated  unit  stresses  are  very  light 
which  is  a  great  advantage  on  a  bad  foundation. 

102.  Flood  Pressures.  The  second,  or  flood  stage,  will  now  be 
investigated.  Here  the  vertical  load  line  N  in  Fig.  75e  is  increased 
by  140  tons,  the  additional  weight  of  water  carried  by  the  arch.  The 
horizontal  water  pressure  PI  is  now  763  tons  and  ]V=1018,  their 
resultant  being  RI.  The  reverse  pressure  due  to  a  depth  of 
12  feet  of  water  is  70  tons,  this  combined  with  RI,  in  Fig.  75a, 
results  in  R%  the  final  resultant.  The  value  of  6  is  35°  15'  which  is 
satisfactory.  As  q  scales  5  feet,  the  unit  stresses  work  out  as 
follows : 


DAMS  AND  WEIRS  131 

At  heel  3.9  tons,  at  center  2.7,  and  at  toe,  4.2  tons. 

The  stress  in  the  arch  under  a  head  of  38  feet  comes  to  11.5 
tons.  Thus  the  stresses  in  stage  2  are  higher  than  is  the  case  with 
stage  1. 

At  the  end  of  a  series  of  these  scallop  arches  near  either  abut- 
ment the  thrust  of  the  arch  resolved  axially  with  the  weir  has  to 
be  met  either  by  tying  the  last  two  arches  by  a  cross  wall  and  rein- 
forcing rods,  or  abutting  the  arch  on  an  abutment  supported  by 
wall  or  a  length  of  solid  dam.  This  design  would,  it  is  considered, 
be  improved  if  the  versed  sine  of  the  arcs  were  made  somewhat 
greater,  as  the  arches  are  too  flat  near  the  crest. 

The  following  remarks  bear  on  the  curvature  of  the  arch  men- 
tioned in  section  101.  When  a  segmental  arch  is  inclined,  the 
spring  line  is  at  a  lower  level  than  the  crown,  consequently  the 
wrater  pressure  is  also  greater  at  that  level.  But  the  thickness 
should  vary  with  the  pressure  which  it  does  not  in  this  case.  This 
proves  the  advisability  of  making  the  circular  curvature  horizontal, 
then  a  section  at  right  angles  to  the  inclined  spring  line  will  be  an 
ellipse,  while  a  horizontal  section  will  be  a  segment  of  a  circle. 
The  reverse  occurs  with  arches  built  in  the  ordinary  way.  There 
appears  to  be  no  practical  difficulty  in  constructing  forms  for  an 
inclined  arch  on  this  principle. 

103.  Big  Bear  Valley  Dam.  Fig.  76  is  a  plan  and  sectional 
elevation  of  the  new  Bear  Valley  reinforced  concrete  multiple 
arch  dam  which  takes  the  place  of  the  old  single  arch  dam  men- 
tioned in  section  83.  The  following  description  is  taken  from 
"Engineering  News",  from  which  Fig  78  is  also  obtained. 

The  new  dam  consists  of  ten  arches  of  30J  feet,  clear  span  at 
top,  abutting  on  eleven  buttresses.  The  total  length  of  the  dam 
is  363  feet  on  the  crest;  its  maximum  height  from  crest  to  base  is 
92  feet  (in  a  pocket  at  the  middle  buttress  only),  although,  as  the 
elevation  in  Fig.  76  shows,  the  average  height  of  the  buttresses 
is  much  less  than  that  figure.  The  water  face  of  the  structure 
and  the  rear  edge  of  the  buttresses  are  given  such  slopes  as  to  bring 
the  resultant  of  the  water-pressure  load  and  that  of  the  structure 
through  the  center  of  the  base  of  the  buttresses  at  the  highest 
portions  of  the  dam,  Fig.  79.  The  slope  for  the  water  face 
up  to  within  14  feet  of  the  top  is  36°  52'  from  the  vertical,  and 


132 


DAMS  AND  WEIRS 


from  that  point  to  the  crest  is  vertical.  The  slope  of  the  down- 
stream edges  of  the  buttresses  is  2  on  1  from  the  bottom  to  the  top, 
the  vertical  top  of  the  face  arches  giving  the  piers  a  top  width  of  10 
feet  from  the  spring  line  to  the  back"  edge.  The  buttresses  are  1.5 
feet  thick  at  the  top  and  increase  in  thickness  with  a  batter  oi 
0.016  feet  per  foot  of  height  or  1  in  60  on  each  side  to  the  base  for 
all  heights.  The  arch  rings  are  12  inches  thick  at  the  top  and 
down  to  the  bend,  from  which  point  they  are  increased  in  thick- 
ness at  the  rate  of  0.014  feet  per  foot  to  the  base,  or  1  in  72.5 


Fig.  77.     View  of  Big  Bear  Valley  Dam  with  Old  Dam  Shown  in  Foreground 


The  arc  of  the  extrados  of  the  arch  ring  is  140°  08'  from  the 
top  to  bottom  the  radius  being  maintained  at  17  feet  and  the  rise 
at  11.22  feet.  The  extrados  is,  therefore,  a  cylindrical  surface 
uniform  throughout,  all  changes  in  dimensions  being  made  on  the 
intrados  of  the  arch.  Thus  at  the  top,  the  radius  of  the  intrados 
is  16  feet,  the  arc  145°  OS',  and  the  rise  11.74  feet.  At  80  feet  from 
the  top,  Fig.  79,  the  thickness  of  the  arch  ring  will  be  2.15  feet, 
the  radius  of  the  intrados  14.85  feet  (the  radius  of  extrados  less 
the  thickness  of  the  wall),  the  arc  140°^  48'  and  the  rise  10.59 


DAMS  AND  WEIRS  133 

feet.     In  all  cases  of  arch-dam  design  the  clear  span,  radius,  and 
rise  of  the  intrados  decrease  from  the  top  downward. 


Strut-tie  members  are  provided  between  the  buttresses  to 
stiffen  and  take  up  any  lateral  thrusts  that  might  be  set  up  by 
seismic  disturbances  or  vibrations,  these  consisting  of  T-beams 


134 


DAMS  AND  WEIRS 


and  supporting  arches  all  tied  together  by  heavy  steel  reinforcement. 
The  T-beams  are  12  inches  thick  and  2.5  feet  wide,  with  a  12- 
inch  stem,  set  on  an  arch  12  inches  square  at  the  crown  and  thick- 
ening to  15  inches  toward  the  springing  lines,  with  two  spandrel 
posts  on  each  side  connecting  the  beam  and  arch,  all  united  into 
one  piece.  There  are  provided  copings  for  the  arches  and  the  tops 
of  the  buttresses  with  9-inch  projections,  making  the  arch  cope 
2.5  feet  wide  and  that  on  top  of  the  buttresses  3  feet  wide.  The 
beam  slab  of  the  top  strut  members  is  built  4  feet  wide  to  serve 
as  an  extra  stiffener,  as  well  as  a  comfortable  footwalk  across  the 
dam.  This  footwalk  is  provided  with  a  cable  railing  on  both  sides 


Scale  10'*  I " 

Fig.  79.     Profile  and  Sections  of  Big  Bear  Valley  Dam 

to  make  it  a  safe  place  upon  which  to  walk.  To  add  to  the  archi- 
tectural effect  of  the  structure,  the  arches  of  the  strut  members 
terminate  in  imposts,  built  as  part  of  the  buttresses.  The  struts 
are  reinforced  with  twisted  steel  rods,  all  being  tied  together  and 
all  being  continuous  through  the  buttress  walls.  The  ends  entering 
the  buttresses  are  attached  to  other  reinforcement  passing  cross- 
wise into  the  buttress  walls,  forming  roots  by  which  the  stresses 
in  the  beams  may  be  transmitted  to  and  distributed  in  the  buttress 
walls.  The  ends  of  the  strut  members  are  all  tied  onto  the  granite 
rock  at  both  ends  of  the  structure  by  hooking  the  reinforcement 
rods  into  drill  holes  in  the  rock.  The  buttresses  are  not  reinforced, 
except  to  be  tied  to  the  arch  rings  and  the  strut  members,  their 


DAMS  AND  WEIRS  135 

shape  and  the  loads  they  are  to  carry  making  reinforcement  super- 
fluous. The  arch  ribs  are  reinforced  with  f-inch  twisted  rods  hori- 
zontally disposed  2  inches  from  the  inner  surface  and  variably 
spaced.  These  rods  were  tied  to  the  rods  protruding  from  the 
buttresses.  For  reinforcing  the  extrados  of  the  arch  ring  ribs  of 
liXl|Xft-inch  angles  were  used,  to  which  "ferro-inclave"  sheets 
were  clipped  and  used  both  as  a  concrete  form  for  the  outer  face 
and  a  base  for  the  plaster  surface. 

104.  Stress  Analysis.  On  Fig.  79  a  rough  stress  analysis 
is  shown  for  80  feet  depth  of  water.  As  will  be  seen  the  resultant 
R  cuts  the  base  just  short  of  the  center  point.  The  value  of  N 
is  estimated  at  4100  tons,  the  area  of  the  base  J>  110X4.2  =  460 

sq.  feet  whence  —  =  ——  -  =  9  tons  nearly,  evenly  distributed  (m  being 


taken  as  unity).     The  stress  on  the  arch,  80  feet  deep,  neglecting 

.        .    RHw     16X80X1     on  .  rp,  .      , 

its  weight  is  —  -  ---  —  —  ——  =  20  tons,  nearly.     Ihis  shows  the 


1         3 
necessity  for  the  reinforcement  provided  to  take  —  or  —  of  this  stress. 


P 
The  tangent  of  ^  =      =          =  . 


This  is  a  large  value,  35  degrees  being  the  usual  limit,  33  degrees 
better.  If  the  arch  thickness  were  doubled,  reinforcement  would  not 
be  necessary  except  near  the  crest  and  the  additional  load  of  about 
320  tons  would  bring  6  down  to  35  degrees.  If  not,  a  greater  inclination 
given  to  the  arch  would  increase  the  load  of  water  on  the  extrados. 
It  is  quite  possible  that  a  thicker  arch  without  reinforcement  would 
be  actually  cheaper.  The  downward  thrust  acting  on  the  arch 
due  to  its  own  weight  is  on  a  different  plane  from  the  arch  thrust. 
Its  effect  is  to  increase  the  unit  stress  to  a  certain  extent,  as  is  also 
the  case  with  the  combination  of  shearing  and  compressive  stresses 
in  the  interior  of  a  dam  as  explained  in  Part  I.  This  increase  can, 
however,  be  neglected.  A  considerable  but  undefined  proportion 
of  the  water  pressure  near  the  base  is  conveyed  to  it  and  not  to 
the  buttresses;  this  will  more  than  compensate  for  any  increase 
due  to  vertical  compression  and  consequently  it  can  be  ignored. 
The  ribs  connecting  the  buttresses  form  an  excellent  provision  for 
stiffening  them  against  buckling  and  vibration  and  are  universally 


136  DAMS  AND  WEIRS 

employed  in  hollow  concrete  dams.     The  buttresses  in  this  instance 
are  not  reinforced. 


HOLLOW  SLAB  BUTTRESS  DAMS 

105.  Description  of  Type.  There  is  a  class  of  dam  and  weir 
similar  in  its  main  principles  to  the  arch  buttress  type  which  is 
believed  to  have  been  first  introduced  by  the  Ambursen  Hydraulic 
Construction  Company  of  Boston.  In  place  of  the  arch  an  inclined 
flat  deck  is  substituted,  which  has  necessarily  to  be  made  of  rein- 
forced concrete.  For  this  reason,  the  deck  slabs  cannot  exceed  a 
moderate  width,  so  numerous  narrow  piers  take  the  place  of  the 
thick  buttresses  in  the  former  type.  A  further  development  is  a 
thin  deck  which  covers  the  downstream  ends  of  the  buttresses  or 
piers,  forming  a  roll  way.  The  enclosed  box  thus  formed  is  occa- 
sionally utilized  as  a  power  house  for  the  installation  of  turbines, 
for  which  purpose  it  is  well  suited. 

The  inclination  given  to  the  flat  deck  is  such  that  the  incidence 
of  the  resultant  (R.F.)  will  fall  as  near  the  center  of  the  base  as  pos- 
sible and  at  the  same  time  regulate  the  inclination  of  the  resultant 
to  an  angle  not  greater  than  that  of  the  angle  of  friction  of  the 
material,  i.e.,  30  degrees  with  the  vertical.  By  this  means  any 
tendency  to  slide  on  the  foundation  is  obviated. 

Ellsworth  Dam  an  Example.  A  good  example  of  this  style  of 
construction  is  given  in  Fig.  80  of  the  Ellsworth  dam  in  Maine.  In 
this  design  the  inclination  of  the  deck  is  45°  or  very  nearly  so;  the 
piers  are  15  feet  centers  with  widened  ends,  so  that  the  clear  span 
of  the  concrete  slabs  is  .9'  1"  at  the  bottom. 

The  calculations  necessary  to  analyze  the  thickness  of  the 
slabs  and  the  steel  reinforcement  at  one  point,  viz,  at  El.  2.5,  will 
now  be  given.  In  this  case  the  pressure  of  water  on  a  strip  of  the 
slab,  one  foot  wide,  the  unsupported  span  of  which  is  9'  1",  is  Hlw. 
Here  H  =  67  feet  and  w  is  ^  ton  per  cubic  foot;  therefore,  W  = 
67X9.1XA  =19  tons.  To  this  must  be  added  the  weight  of  the 
slab.  As  this  latter  lies  at  an  angle  with  the  horizontal  its  weight 
is  partly  carried  by  the  base  and  is  not  entirely  supported  by  the 
piers.  The  diagram  in  Fig.  80c  is  the  triangle  of  forces.  The  weight 
of  slab  w  is  resolved  in  two  directions,  a  and  6,  respectively,  parallel 


DAMS  AND  WEIRS 


137 


and  normal  to  face  of  slab.     The  angle  being  45  degrees  a-b=  W- 

V2* 
Consequently  the  thickness,  37  inches,  can  be  considered  as  reduced 

-  to  |4  =  2.2  feet.     The 


OOOI 


SNO-L 
OOf 


J-" 


portion  of  the  weight  of 
the  slab  carried  to  the 
piers  will,  therefore,  be 

9.1  X2.2X~  =  1.5  tons, 

tKe  weight  of  the  con- 
crete being  assumed  at 
the  usual  value  of  150 
pounds  per  cubic  foot. 
g  The  total  distributed 
E  load  in  the  strip  will  then 
|  be  19 +  1.5  =  20.5  tons. 
£  Now  the  moment  of 

£  stress  on  a  uniformly 
of  loaded  beam  with  free 
P  Wl 

CD  -I  •  \V    I  ^  r 

g    ends   is  — ,   or   M  = 

Cu  O 


1    20.5X109 

s"       8 

g 

ft    tons. 


279    inch- 


This  moment  must 
be  equaled  by  that  of 
the  resistance  of  the 
concrete  slab. 

106.  Formulas  for 
Reinforced  Concrete. 
For  the  purpose  of 
showing  the  calcula- 
tions in  detail,  some 
leading  formulas  con- 
nected with  reinforced 
beams  and  slabs  will 
now  be  exhibited : 


138  DAMS  AND  WEIRS 

bd?=^-8.  (26) 

fsPJ 

or,  approximately,  bd?  =  jj-1  (26a) 

8JsP 

(27) 


or,  approximately,  W  =  —  c  (27a) 

6jc 

From  these  are  found  d,  the  required  thickness  of  a  slab  up  to 
centroid  of  steel,  or  MS+MC  the  bending  moments,  in  which  b  is 
width  of  beam  in  inches;  d  depth  of  centroid  of  steel  below  top 
of  beam;  Mc  and  Ms  symbolize  the  moments  of  resistance  of  the 
concrete  and  steel,  respectively;  /s  safe  unit  fiber  stress  in  steel, 
12,000  to  16,000  lb.,  or  6  to  8  tons  per  square  inch;/c  safe  extreme 

stress  in  concrete  500,  600,  or  650  lb.,  or  .25,  .3,  or  .325  ton  per  square 

^ 
inch;  p  steel  ratio,  or  — 

bd 

Ideal  steel  ratio  p  =          —  r  (28) 

A  area  of  cross-section  of  steel  ;  k  ratio  of  depth  of  neutral  axis  below 
top  to  depth  of  beam 

pn  (29) 


j  ratio  of  arm  of  resisting  couple  to  d 

j  =(!-!*)  (30) 

Tji 

n  ratio  ~,  Es  and  Ec  being  the  moduli  of  elasticity,  ordinary  values 

EC 

12  to  15;  r  ratio  —  .     As  p  =  rr>  when  reinforced  slabs  are  analyzed, 
fc  bd 

formulas  (26)  and  (27)  can  be  transposed  as  below. 

From  (26)    Ms=fsAjd  (31) 

From  (26a)  Ms  =  lfsAd    Approximate  (3  la) 

From  (27)     Mc  =  ifckjbd*  (32) 


From  (27a)  Me=-         Approximate  (32a) 

6 

In  the  case  under  review  the  reinforcement  consists  of  three 
one  inch  square  steel  rods  in  each  foot  width  of  the  slab.    Using  the 


DAMS  AND  WEIRS  139 

approximate    formulas    (3  la)    and    (32a),  /s  =  8  tons,  /c  =  .3   ton, 
d  =  35  inches  and  6  =  12  inches;  then 

-X35  =  735  inch-tons 


Mc  =  2-  X  4-  X  420  X  35  =  735  inch-tons 
10     6 

the  results  being  identical.  As  already  noted  the  moment  of  stress 
is  but  279  inch-tons.  The  end  shear  may  have  governed  the  thick- 
ness. Testing  for  shear  the  load  on  a  12-inch  strip  of  slab  is  20.5 
tons  of  which  one-half  is  supported  at  each  end.  Allowing  50  lb.,  or 
.025  ton,  as  a  safe  stress,  the  area  of  concrete  required  is  10.25  -f-  .025 
=  410  square  inches  the  actual  area  being  37  X  12  =  444  square  inches. 

107.  Steel  in  Fore  Slope.  The  reinforcement  of  the  fore  slope 
is  more  a  matter  of  judgment  than  of  calculation,  this  deck  having 
hardly  any  weight  to  support,  as  the  falling  wTater  will  shoot  clear 
of  it.  The  piers  are  not  reinforced  at  all,  nor  is  it  necessary,  as  the 
stresses  are  all  compressive  and  the  inclination  of  the  upstream  deck 
is  such  that  the  resultant  pressure  makes  an  angle  with  the  vertical 
not  greater  than  that  of  friction,  i.e.,  30  degrees.  Fig.  80a  is  a  force 
diagram  of  the  resultant  forces  acting  on  the  base  at  EL  0.00.  The 
total  weight  of  a  15-foot  bay  is  estimated  at  783  tons  while  that  of  P, 
the  trapezoid  of  water  pressure,  is  1700  tons.  The  force  line  P  in  Fig. 
80  drawn  through  the  e.g.  of  the  water  pressure  area  intersects  the 
vertical  force  W  below  the  base  line.  From  this  intersection  R  is 
drawn  upward  parallel  to  its  reciprocal  in  the  force  polygon,  cutting 
the  base  at  a  point  some  9  feet  distant  from  the  center  point. 

The  maximum  stress  will  occur  at  the  heel  of  the  base.     A  = 

107X2  =  214  sq.  ft.;  -^  =  ^==9.34  tons;  q  being  9  ft.,  m  =  10^54 
A      z!4  107 

=  1  .5  and  s  =  9.34  X  1  .5  =  14  tons.     Formula  (9)  ,  Part  I.    The  hori- 

zontal component  of  P  =  1200  tons.     The  base  being  2  ft.  wide, 

1200 


2X107 


=  5.6  tons;   therefore  by  formula  (10),  Part  I, 


V49+31.4  =  16.5  tons,  a  decidedly  high  value.  The  usual  limit  to 
shearing  stress  is  100  lb.  per  sq.  inch,  equivalent  to  7.2  tons  per 
sq.  ft.,  reinforcement  is  therefore  not  necessary  and  is  not  provided. 
There  appears  to  be  no  reason  why  a  steeper  slope  should  not 
have  been  given  to  the  deck  so  as  to  bring  the  center  of  pressure  up 


140  DAMS  AND  WEIRS 

to  the  center  of  the  base  and  thus  reduce  the  unit  stress.  Possibly  a 
higher  river  stage  has  been  allowed  for.  The  position  of  W  as  well 
as  the  weight  of  the  structure  were  obtained  from  the  section  given 
in  Schuyler's  Reservoirs.  Fig.  80  is  of  the  so-termed  "Curtain" 
type  of  dam.  The  "Half  Apron"  type,  Fig.  82c,  is  sometimes  used  for 
overfalls,  the  main  section  of  Fig.  82  illustrating  the  "Bulkhead"  type. 
108.  Slab  Deck  Compared  with  Arch  Deck  Dam.  The  Ambur- 
sen  dam,  wherever  the  interior  space  is  not  required  for  installa- 
tion of  turbines,  is  undoubtedly  a  more  expensive  construction  than 
the  multiple  arch  type.  This  fact  has  at  last  been  recognized  and  in 
one  of  the  latest  dams  erected,  scallop  arches  were  substituted  for 
the  flat  deck,  thus  obviating  the  expense  of  reinforcement.  By 


Fig.  81.     Section  of  Arch  with  30-Foot  Span 

increasing  the  width  of  the  spans,  the  piers,  being  thicker  in  like 
proportion,  will  be  in  much  better  position  for  resisting  compressive 
stress,  as  a  thick  column  can  stand  a  greater  unit  stress  than  a  thin 
one.  Another  point  in  favor  of  the  arch  is  that  the  effective  length 
of  the  base  of  the  piers  extends  practically  to  the  crown  of  the  arch. 
The  arch  itself  need  not  be  as  thick  as  the  slab.  Owing  to  the  liquid 
radial  pressure  to  which  it  is  subjected  it  is  in  a  permanent  state  of 
compression  and  does  not  require  any  reinforcement  except  possibly 
at  the  top  of  the  dam.  Here  the  arch  is  generally  widened,  as  in 
the  case  of  the  Ogden  dam,  Fig.  73,  and  thus  greatly  stiffened  at  the 
point  where  temperature  variations  might  develop  unforeseen  stresses. 
Fig.  81  is  a  sketch  illustrative  of  the  saving  in  material  afforded 
by  doubling  the  spans  from  15  to  30  feet  and  conversion  to  multiple 
arch  type.  The  radius  of  the  extrados  of  the  arches  is  18.5  ft.  H 
is  67  at  elevation  2.50  and  w=  ^  ton;  hence  the  thickness  of  the 
arch  by  formula  (21)  (si  being  taken  as  15  tons),  will  be 


DAMS  AND  WEIRS 


141 


It  is  thus  actually  thinner  than  the  reinforced  slab  of  one-half 
the  span,  or  15  feet.  The  greater  length  of  the  arch  ring  over  that 
of  the  straight  slab  is  thus  more  than  compensated.  The  area  of 
the  arch,  counting  from  the  center  of  the  pier,  is  35X2.6  =  91  square 
feet,  that  of  the  slab  is  30X3.1  =93,  that  of  the  bracketing  at  junc- 
tion with  the  piers,  13,  giving  a  total  of  106  square  feet.  The  saving 


llway  Cross  Section  of  Spillway 
™*  (C) 


Fig.  82.     Profile  and  Detailed  Sections  of  Guayabal  Dam,  Porto  Rico 

due  to  decreased  length  of  the  piers  is  25  square  feet.  Thus  in  the 
lower  part  of  the  dam  over  40  cubic  feet  per  30'  bay  per  foot  in  height 
of  concrete  is  saved,  also  all  the  steel  reinforcement.  If  a  roll  way 
is  considered  necessary  in  the  weir,  the  deck  could  be  formed  by  a 
thin  reinforced  concrete  screen  supported  on  I-beams  stretching 
across  between  the  piers. 

109.  Guayabal  Dam.  Fig.  82  is  a  section  of  the  Guayabal 
dam  recently  constructed  in  Porto  Rico,  its  height  is  127  feet  and 
it  is  on  a  rock  foundation.  The  following  are  the  conditions  govern- 


142  DAMS  AND  WEIRS 

ing  the  design;  maximum  pressure  on  foundation  10  tons  per  square 
foot;  compression  in  buttresses  300  pounds  per  square  inch  or  21.6 
tons  per  square  foot;  shear  in  buttresses  100  pounds  per  square 
inch,  or  7.2  tons  per  square  foot;  shear  in  deck  slabs  60  pounds,  or 
.03  ton  per  square  inch;  fc  for  deck  slabs  600  pounds  or  .3  ton  per 
square  inch;  fs  for  deck  slabs  14,000  pounds,  or  7  tons  per  square 
inch. 

The  concrete  in  the  slabs  is  in  the  proportion  of  1:2:4,  in  the 

E  f 

buttresses  1:3:6;    n  =  —  s  is  taken  as  15  and  r=V=23.3.     Thedeck 

&c  fc 

slab  is  55  inches  thick  at  EL  224,  d  is  taken  as  53,  allowing  2  inches 
for  covering  the  steel,  bd  or  the  area  of  the  section  one  foot  wide  = 
53  X  12  =  636  square  inches.  Now  A  the  area  of  the  steel  =pbd.  By 


formula  (28).  P--  =  -01°44'  hence 


required  area  of  steel  will  be  636  X.  01044  =  6.64  square  inches, 
provided  d  is  of  the  correct  value.  The  calculation  wrill  now  be 
made  [for  the  thickness  of  the  slab  which  is  actually  55  inches. 
The  load  on  a  strip  12  inches  wide  is 

109X13 

Water  pressure  -  -  —  =  44.3  tons     . 
oZ 

To  this  must   be  added  a  portion  of  the   weight  of  the   slab 

(13X55\     f  3  \ 
—  —  —  1  X  (  —  j  =  4.5   tons.     Of   this 

-^-  =  3.2  tons  must  be  added  to  the  44.3  tons  above,  44.3+3.2  =  47.5 

V2 

rru    u     j-  IT-    WL    47.5X13X12      no_  . 

tons.     The  bending  moment  M  is  —  —  =  -  —  =  927  mch- 

8  8 

tons.  The  depth  of  the  slab  can  be  estimated  by  using  formulas 
(26)  or  (27)  or  the  approximate  ones  (26a)  and  (27  a).  For  the 
purpose  of  illustration,  all  four  will  be  worked  out.  First  the  values 
of  k  and  j  will  be  found  by  formulas  (29)  and  (30). 

&  =  V.313+.0245-.156  =  .582-.156  =  0. 


By  formula  (26),  ,. 


DAMS  AND  WEIRS  143 


d  =  V1234  =  35.07  inches 

9  If  Q97 

By  fonnula  (27),  *-. 


d  =  1406  =  37.5  inches 
Now  the  approximate  formulas  will  be  used.     By  (26a) 


8X927  =  1210 


7X12X7X.0104     1.53 
d  =  Vl210  =  34.8  inches 
by  (27a) 

* 


d  =  Vl542  =  39.3  inches 

The  approximate  formulas  (26a)  and  (27a)  give  higher  results  than 
(26)  and  (27)  .  The  result  to  select  is  37.5  inches,  formula  (27)  ,  which  is 
higher  than  by  (26)  .  The  depth  of  beam  would  then  be  40  or  41  inches. 
It  is  actually  55.  This  discrepancy  may  be  due  to  the  water  having 
been  given  a  s.g.  in  excess  of  unity,  owing  to  the  presence  of  mud 
in  suspension,  say  of  1.3  or  1.5,  or  shear  is  the  criterion. 

The  corresponding  steel  area  will  be  A=pbd  =  .  0104X12X37.5 
=  4.7  square  inches.  1  &  "  round  rods  spaced  3  inches  would  answer. 
With  regard  to  direct  shear  on  the  slab,  W  as  before  =47.5  tons  of 
which  half  acts  at  each  pier,  viz,  23.7  tons.  The  safe  resistance  is 

bd  X  Ss  =  12  X  55  X  .03  =  20  tons,  nearly.     The  shear  =  Jp'      =  .036 

\Z  Xoo 

ton  =  72  pounds  per  square  inch.  This  figure  exceeds  the  limit  of 
60  pounds.  The  deficiency  is  made  up  by  adding  the  shear  of  the 
steel  rods.  The  sectional  area  of  this  reinforcement  is  4.7  square 
inches  the  safe  shearing  of  which  is  over  20  tons.  These  rods  are 
usually  turned  up  at  their  ends  in  order  to  care  for  the  shear. 

Shear  in  Buttresses.  With  regard  to  shear  in  the  buttresses,  the 
horizontal  component  of  the  water  pressure  as  marked  on  the  force 
diagram  is  3400  tons.  The  area  of  the  base  of  the  buttress  at 

El.  224  is  138X3.2  =  441.6,  the  shearing  stress  or  ss  then  =^Lg 

441.6 

tons  per  square  foot,  nearly.  The  allowable  stress  being  only  7.2 
tons  the  difference  will  have  to  be  made  good  by  reinforcing  rods, 
of  which  two  of  f-inch  diameter  would  suffice. 


144 


DAMS  AND  WEIRS 


Now  with  regard  to  compressive  stresses  on  the  buttresses  the 
graphical  working  shows  that  the  resultant  R  strikes  the  base  at  EL 

224  almost  exactly  at  the  center,  the  angle 
6  also  is  30  degrees.  The  value  of  N  is 
5650  tons;  si  the  mean  and  s  the  maximum 

N 

stress  will  both  equal—;  and  A, the  area 
A 

of  the  base,  equals  138X3.2  =  442  sq.  ft.; 

r*  c*  r  d 

therefore,  s  =  ——  =  12.78  tons.  The  com- 
pression on  the  foundation  itself,  which 
is  4  feet  lower  will  not  be  any  less  for, 
although  the  base  width  is  greater,  N  as 
well  as  P  are  also  increased.  Thus  the 
pressure  on  the  foundation  is  in  excess 
of  the  limit  and  widening  to  a  further 
extent  is  required. 

The  maximum  internal  stress  c,  in 
the  buttress  at  El.  224,  will  be  by  formula 


(10),  Part  I,  is+^l— +s,2.  Here  s  =  12.8 
and  ss  as  we  have  seen  is  8  tons,  therefore, 

c  =  6.4+J^+64  =  16.6  tons.   The  limit 
\    4 

compression  in  the  buttress  is  300  pounds 
per  square  inch,  or  21.6  tons  per  square 
foot. 

In  the  bulkhead  portion  of  the  dam, 
shown  in  Fig.  82b,  every  pier  is  run  up 
14  inches  thick  through  the  deck  to  form 
a  support  for  a  highway  bridge,  the  spans 
of  which  are  therefore  16  feet  10  inches 
in  the  clear;  the  roadway  is  carried  on 
slabs  which  are  supported  by  arches  of 
reinforced  concrete.  The  buttresses  are 
laterally  supported  by  several  double  rein- 
forced sway  beams,  16"X14",  and  below  the  crest  a  through  road- 
way is  provided.  The  spillway  section  is  shown  on  Fig.  82c.  The 


DAMS  AND  WEIRS 


145 


146  DAMS  AND  WEIRS 

ground  level  is  here  on  a  high  bench  at  El.  295.  The  crest  being 
El.  325,  the  fall  is  30  feet.  The  spillway  is  of  the  "half  apron  type". 
The  roadway  here  is  carried  on  four  reinforced  concrete  girders,  a 
very  neat  construction;  the  piers  are  run  up  every  alternate  span 
and  are  therefore  at  36-foot  centers;  they  are  beveled  on  both  faces 
to  reduce  end  contraction.  The  spillway  will  pass  70,000  second- 
feet;  its  length  is  775  feet. 

The  bulkhead  section  of  the  dam  (see  also  Fig.  83)  has  51  spans 
of  18-foot  centers,  total  length  918  feet;  that  of  the  spillway  consists 
of  21  spans  of  36-foot  centers.  The  whole  length  is  1674  feet.  The 
depth  of  the  tail  water  is  not  known,  it  would  probably  be  about 
20  feet  and  its  effect  would  be  but  trifling.  This  is  one  of  the  largest 
hollow  dams  ever  constructed.  The  arrangement  of  the  haunches 
or  corbels  of  the  buttresses  is  a  better  one  than  that  in  the  older 
work  of  Fig.  80. 

110.  Bassano  Dam.  Another  important  work  is  the  Bassano 
dam  illustrated  in  Figs.  84  and  85.  This  is  an  overfall  dam  built  over 
the  Bow  River  at  the  head  of  the  eastern  section  of  the  Canadian 
Pacific  Railway  Company's  irrigation  canal  and  is  estimated  to  pass 
100,000  second-feet  of  water  at  a  depth  of  14  feet.  Though  not  so 
high  nor  so  long  as  the  Guayabal  dam  it  presents  several  features  of 
interest.  First  its  foundations  are  on  a  thick  blanket  of  clay  some 
twelve  feet  deep  which  overlies  boulders  and  gravel.  •  This  material 
is  very  hard  blue  clay  of  excellent  quality.  The  great  advantage  of 
this  formation,  which  extends  over  1000  feet  upstream  from  the  work, 
is  that  it  precludes  all  uplift,  or  very  nearly  so,  consequently  no 
special  precautions  have  to  be  adopted,  such  as  a  long  apron  to 
ensure  length  of  percolation,  as  would  be  necessary  in  case  of  a 
foundation  composed  of  porous  and  loose  materials.  It  has  also 
disadvantages.  The  allowable  pressure  on  the  clay  is  limited  to 
2J  tons  per  square  foot.  This  influences  the  design  necessitating 
a  wide  spread  to  the  buttresses,  laterally  as  well  as  longitudinally. 
The  whole  of  the  dam  is  an  overfall  and  the  general  arrangements 
are  very  similar  to  those  prevailing  at  Guayabal.  The  hearth  or 
horizontal  fore  apron,  a  provision  not  necessary  in  the  last  example, 
is  at  EL  2512.  The  crest  is  at  2549.6  a  height  of  37.6  feet  above 
the  apron  and  corresponds  with  the  level  of  the  canal  intake  floor. 
Water  is  held  up  to  eleven  feet  above  crest  level  by  draw  gates 


DAMS  AND  WEIRS 


147 


148  DAMS  AND  WEIRS 

eleven  feet  high,  and  this  full  supply  level  is  three  feet  below  that  of 
the  estimated  afflux,  which  is  fourteen  feet  above  the  crest. 

For  overturning  moment  the  water-pressure  area  will  be  a 
truncated  triangle  with  its  apex  at  afflux  level  plus  the  height  h  or 

v2 
1.5  • — •  to  allow  for  velocity  of  approach,  as  explained  in  section 

57,  Part  I.     This,  in  the  Bow  River  with  a  steep  boulder  bed  will  be 

144 
about  12  feet  per  second;  h  therefore  will  equal  LSX-rj  =3.4  feet  and 

OT: 

the  apex  of  the  truncated  triangle  will  be  at  a  point  14+3.4  =  17.4 
feet  above  the  crest  level.  The  depth  of  the  tail  water  at  full  flood 

is  not  known,  the  ratio  —  with  a  steep  bed  slope  will  not  be  under  .5, 

consequently  with  d  =  14,  D  will  have  a  value  of  about  25  to  28  feet, 
d  being  depth  over  crest  and  D  that  of  tail  water.  The  overturning 
moments  direct  and  reverse  can  be  represented  by  the  cubes  of  the 
depths  up-  and  downstream  and  the  unbalanced  moment  by  their 
difference.  The  upstream  head  is  37.0+14+3.4  =  55  feet  and  the 
downstream  head  say  25  feet.  Their  cubes  are  166,375  and  15,625 
the  difference  being  150,750,  thus  the  reverse  pressure  will  not  have 
much  effect  in  assisting  the  stability  of  the  structure.  The  cor- 
responding representative  moment  when  water  is  held  up  to  11  ft. 
above  crest  will  be  493  =  117,649,  supposing  the  tail  channel  empty. 
This  quantity  is  less  than  the  150,750  previously  stated,  consequently 
the  afflux  level  is  that  which  has  to  be  considered  when  estimating 
the  overturning  moment.  In  the  case  of  direct  water  pressure  on 
the  deck  slabs,  the  acting  head  at  full  flood  will  be  the  difference  of 
the  flood  level  up-  and  downstream,  which  is  30  feet,  as  the  tail 
water  is  allowed  access  to  the  rear  of  the  deck  slabs.  This  is  less 
than  the  head,  49  feet,  which  exists  when  the  gates  are  closed  and 
water  is  held  up  to  canal  full  supply,  i.e.,  to  EL  2560.6,  consequently 
the  head  that  has  to  be  considered  is  that  at  this  latter  stage. 

Analysis  of  Pressures  on  Bassano  Dam.  With  this  data  the 
design  can  be  analyzed,  the  procedure  being  identical  with  that 
explained  in  the  last  example,  excepting  that  the  reverse  pressure 
might  be  taken  into  account  as  it  will  modify  the  direction  and 
incidence  of  72  in  a  favorable  sense  though  not  to  any  great  extent. 
The  limit  stresses  are  those  given  in  the  last  example  with  the  fol- 


DAMS  AND  WEIRS  149 

lowing  additions:    Footings,  compression  in  bending,  600  pounds 
per  square  inch,  shear,  75  pounds  per  square  inch. 

Some  explanation  will  now  be  given  of  the  method  of  calcu- 
lation of  the  footing  to  the  buttresses  and  the  Guayabal  dam  will 
be  referred  to,  as  the  pressures  on  the  base  of  the  buttresses  are 
known  quantities.  In  section  109  the  value  of  N  is  5650  tons  and 

-  =41  tons,  nearly.     This  is  the  unit  pressure  per  foot  run 
o       138 

on  the  base  of  the  buttress.     Supposing  the  limit  pressure  on  the 
foundation  was  fixed  at  3  tons  per  square  foot,  then  the  requisite 

41 

base  width  of  the  footing  would  be  —  =  13.7  feet.     The  footing  con- 

o 

sists  of  two  cantilevers  attached  to  the  stem  of  the  buttress.     The 
bending  moment  M  at  the  junction  with  the  buttress  of  a  strip  1 

Wl 
foot  wide  will  be  —  .     The  buttress  being  3.2  feet  wide  the  pro- 

Zi 

13.7—3.2 

jecting  length  of  footing  on  each  side  will  be  —  '—  —  —  =  5.25  feet. 

The  reaction  on  a  strip  one  foot  wide  will  be  5.25X3X1  = 
15.75  tons.  The  moment  in  inch-tons  about  the  edge  of  the  section 

.„    ,       12WI    15.75X5.25X12          ,    . 

of    the    buttress    will    be   •          =  -  -  —       —  =  498   inch-tons. 

—  2, 


According  to  formula   (27a),   W=-.     :.  d  =  =  V83Q  = 

jc 


28.8  inches.  Then  6d  =  28.8X12  =  346  and  A  the  area  of  the  steel 
at  the  base  will  be  p6d=.  0104X346  =  3.61  inches,  this  in  a  12-inch 
wide  strip  will  take  li-inch  bars  4  inches  apart.  When  the  weight 
on  the  buttress  is  considerable  the  depth  of  footing  slab  thus  esti- 
mated becomes  too  great  for  convenience.  In  such  cases,  as  in  Fig. 
85,  the  beam  will  require  reinforcement  in  compression  at  the  top. 
This  complicates  the  calculation  and  cases  of  double  reinforcement 
are  best  worked  out  by  means  of  tables  prepared  for  the  purpose. 
The  footings  shown  in  Fig.  85,  were  thus  double  reinforced,  in  fact 
through  bars  were  inserted  at  each  step,  the  lower  being  in  tension 
the  upper  ones  in  compression.  The  lower  bars  were  continuous 
right  through  the  base  of  the  dam.  This  reinforcement  of  the 
footing  is  not  shown  on  the  blue  print  from  which  Fig.  85  is  derived. 
111.  Pressure  on  Foundation  Foredeck.  A  great  many 


150  DAMS  AND  WEIRS 

Ambursen  dams  have  been  constructed  on  river  beds  composed 
of  boulders  and  gravel,  which  require  a  pressure  limit  of  about 
4  tons  per  square  foot.  This  can  always  be  arranged  for  by  widen- 
ing the  footing  of  the  pier  buttresses,  the  same  can  of  course  be 
done  with  arch  buttressed  dams.  The  base  of  the  arch  itself  can 
be  stepped  out  in  a  similar  manner.  In  the  Bassano  dam  the 
sloping  fore  deck  is  unusually  thick  and  is  heavily  reinforced  in 
addition;  this  is  done  with  the  idea  of  strengthening  the  structure 
against  shock  from  ice,  as  well  as  from  the  falling  water,  and  with 
the  further  idea  of  assisting  the  buttresses  in  carrying  the  heavy 
load  of  the  piers  and  superstructure.  It  is  doubtful  if  any  calcu- 
lations can  well  be  made  for  this;  it  is  a  matter  more  of  judgment 
than  of  estimation. 

Buttresses.  As  with  the  Guayabal  spillway,  every  alternate 
buttress  is  run  up  to  form  the  piers  of  the  superstructure,  which  latter 
consists  of  a  through  bridge  which  carries  the  lift  gear  for  manipu- 
lating the  draw  gates.  The  so-termed  blind  buttresses — that  is, 
those  that  do  not  carry  a  pier — are  of  thinner  section  and  are  appar- 
ently not  reinforced.  Both  kinds  of  buttresses  have  cross-bracing 
as  shown  on  the  profile.  In  hollow  dams  the  location  of  the  center 
of  pressure  moves  with  the  rise  of  water  from  the  heel  toward  the 
center  within  the  upstream  half  of  the  middle  third.  In  solid  dams, 
on  the  other  hand,  the  movement  is  along  the  whole  of  the  middle 
third  division,  consequently  in  hollow  dams  there  is  no  tendency 
to  turn  about  the  toe  as  with  solid  dams,  rather  the  reverse,  namely, 
to  upset  backward.  This  latter  tendency  must  cause  tension  in 
the  buttresses  which  the  cross-bracing  is  intended  to  care  for. 

Baffles.  As  noted  already  in  section  66,  baffles  have  been  built 
on  the  curved  bucket  with  the  object  of  neutralizing  this  mischiev- 
ous arrangement  which  it  is  hoped  will  soon  Lbecome  as  obsolete  in 
western  practice  as  has  long  been  the  case  in  the  East. 

Hearth  and  Anchored  Apron.  The  dam  is  provided  with  a  solid 
horizontal  fore  apron  or  hearth  76  feet  long  and  beyond  this  the 
device  of  an  anchored  floating  apron  of  timber  30  feet  in  length  has 
been  added.  The  apron  is  undoubtedly  too  short  and  should  have 
been  made  100  feet  or  2  (H+d)  in  length,  with  cribbed  riprap 
below  it.  The  wooden  sheet  piling  in  the  rear  of  the  work  is  con- 
sidered to  be  worse  than  useless;  it  merely  breaks  up  the  good  clay 


DAMS  AND  WEIRS  151 

blanket  by  cutting  it  in  two.  A  wide  solid  curtain  of  concrete,  not 
so  deep  as  to  penetrate  the  clay  blanket,  would  have  been  a  superior 
arrangement.  The  inclined  piling  below  the  bucket  is  provided  to 
guard  against  sliding.  This  dam  is  provided  with  a  number  of  sluice 
openings.  Their  capacity  is  such  that  one  half  will  pass  ordinary 
floods,  allowing  the  other  half  of  the  dam  to  be  cut  off  from  the  river 
by  sheet  piling  during  construction.  On  completion  of  the  work 
these  sluices  were  all  closed  from  the  inside  by  slabs  of  concrete 
deposited  in  position. 

SUBMERGED  WEIRS  FOUNDED  ON  SAND 

1 12.  Description  of  Type.  There  is  a  certain  type  of  drowned 
or  submerged  diversion  weir  which  is  built  across  wide  rivers  or 
streams  whose  beds  are  composed  of  sand  of  such  depth  that  a  solid 
foundation  on  clay  is  an  impossibility.  Consequently,  the  weir  has 
to  be  founded  on  nothing  better  than  the  surface  of  the  river  bed, 
with  perhaps  a  few  lines  of  hollow  curtain  walls  as  an  adjunct.  Of 
this  class  of  weir  but  one  is  believed  to  have  been  constructed  in 
the  United  States,  viz,  the  Laguna  weir  over  the  Colorado  River 
at  the  head  of  the  Yuma  irrigation  canals. 

This  type  originated  in  India  and  in  that  country  are  found 
numerous  examples  of  weirs  successfully  constructed  across  very 
large  rivers  of  immense  flood  discharge.  For  instance,  the  Goda- 
veri  River  in  Southern  India  has  a  flood  discharge  of  1,200,000 
second-feet  and  the  weirs  across  it  are  nearly  2|  miles  in  length. 
Not  only  is  the  length  great,  but  as  will  be  seen,  the  width  has  to  be 
very  considerable.  The  Okhla  weir,  Figs.  101  and  102,  situated  on 
the  Jumna  below  the  historic  city  of  Delhi  is  250  feet  wide  and 
f  mile  long.  The  height  of  these  submerged  weirs  is  seldom  over 
12  feet,  their  role  being  purely  diversion,  not  storage.  No  doubt 
more  of  this  type  of  low  diversion  weirs  will  in  the  future  have  to  be 
constructed  in  the  United  States  or  in  Mexico,  so  that  a  knowledge 
of  the  subject  is  a  necessity  for  the  irrigation  engineer. 

Principles  of  Design.  The  principles  underlying  the  successful 
design  of  these  works  are  a  comparatively  recent  discovery.  Designs 
were  formerly  made  on  no  fixed  principles,  being  but  more  or  less 
modified  copies  of  older  works.  Fortunately  some  of  these  works 
failed,  and  it  is  from  the  practical  experience  thus  gained  that  a 


152 


DAMS  AND  WEIRS 


knowledge  of  the  hydraulic  principles  involved  has  at  last  been 
acquired. 

A  weir  built  on  sand  is  exposed  not  only  to  the  destructive 
influences  of  a  large  river  in  high  flood  which  completely  submerges 
it,  but  its  foundation  being  sand,  is  liable  to  be  undermined  and 
worked  out  by  the  very  small  currents  forced  through  the  under- 
lying sand  by  the  pressure  of  the  water  held  up  in  its  rear.  In  spite 
of  these  apparent  difficulties,  it  is  quite  practicable  to  design  a  work 
of  such  outline  as  will  successfully  resist  all  these  disintegrating 
influences,  and  remain  as  solid  and  permanent  a  structure  as  one 
founded  on  bed  rock. 

113.  Laws  of  Hydraulic  Flow.  The  principle  which  underlies 
the  action  of  water  in  a  porous  stratum  of  sand  over  which  a  heavy 
impervious  weight  is  imposed  is  analogous  to  that  which  obtains  in 


Head  Water  /f 


Fig.  86.     Diagram  Showing  Action  of  Water  Pipe  Leading  Out  of  Reservoir 

a  pipe  under  pressure.  Fig.  86  exemplifies  the  case  with  regard  to 
a  pipe  line  EC,  leading  out  of  a  reservoir.  The  acting  head  (H) 
is  the  difference  of  levels  between  A\  a  point  somewhat  lower  than 
A,  the  actual  summit  level  and  C  the  level  of  the  tail  water  beyond 
the  outlet  of  the  pipe.  The  water  having  a  free  outlet  at  C,  the 
line  AiC  is  the  hydraulic  gradient  or  grade  line.  The  hydrostatic 
pressure  in  the  pipe  at  any  point  is  measured  by  vertical  ordinates 
drawn  from  the  center  of  the  pipe  to  the  grade  line  A  \C.  The  uni- 
form velocity  of  the  water  in  the  pipe  is  dependent  directly  on  the 
head  and  inversely  on  the  f rictional  resistance  of  the  sides  of  the  pipe, 
that  is,  on  its  length.  This  supposes  the  pipe  to  be  straight,  or 
nearly  so. 

114.    Percolation  beneath   Dam.    We  will  now  consider  the 
case  of  an  earthen  embankment  thrown  across  the  sandy  bed  of  a 


DAMS  AND  WEIRS  153 

stream,  Fig.  87.  The  pressure  of  the  impounded  water  will  natu- 
rally cause  leakage  beneath  the  impervious  earthen  base.  With  a 
low  depth  of  water  impounded  it  may  well  be  understood  that  such 
leakage  might  be  harmless;  that  is,  the  velocity  of  the  percolating 
under  current  would  be  insufficient  to  wash  out  the  particles  of 
sand  composing  the  foundation  of  the  dam0  When,  however,  the 
head  is  increased  beyond  a  safe  limit,  the  so-termed  piping  action 
will  take  place  and  continue  until  the  dam  is  completely  undermined. 
115.  Governing  Factor  for  Stability.  The  main  governing 
factor  in  the  stability  of  the  sand  foundation  is  evidently  not  the 
superimposed  weight  of  the  dam,  as  the  sand  is  incompressible; 
although  a  load  in  excess  of  the  hydraulic  pressure  must  exercise 
a  certain  though  possibly  undefined  salutary  effect  in  delaying  the 
disintegration  of  the  substratum.  However  this  may  be,  it  is  the 


y- bp! 
I  iL_4 


Foinl  of£gress  of  Per  eolation  •- 
Fig.  87.     Diagram  Showing  Effect  of  Percolation  under  Earthen  Embankment  across  Stream 

enforced  length  of  percolation,  or  travel  of  the  under  current,  that 
is  now  recognized  to  be  the  real  determining  influence. 

In  the  case  of  a  pipe,  the  induced  velocity  is  inversely  propor- 
tional to  the  length.  In  the  case  under  consideration,  the  hydraulic 
condition  being  practically  identical  with  that  in  a  pipe,  it  is  the 
enforced  percolation  through  the  sand,  and  the  resulting  friction 
against  its  particles  as  the  water  forces  its  way  through,  that  effects 
the  reduction  of  the  velocity  of  the  undercurrent,  and  this  frictional 
resistance  is  directly  proportional  to  the  length  of  passage.  In  the 
case  of  Fig.  87,  the  length  of  enforced  percolation  is  clearly  that  of 
the  impervious  base  of  the  earthen  dam.  The  moment  this  obstruc- 
tion is  passed  the  water  is  free  to  rise  out  of  the  sand  and  the  hydro- 
static pressure  ceases. 

116.  Coefficient  of  Percolation.  This  length  of  enforced  per- 
colation or  travel,  which  will  be  symbolized  by  L,  must  be  some 


154 


DAMS  AND  WEIRS 


multiple  of  the  head  //,  and  if  reliable  safe  values  for  this  factor  can 
be  found,  suitable  to  particular  classes  of  sand,  we  shall  be  enabled 
to  design  any  work  on  a  sand  foundation,  with  perfect  confidence  in 
its  stability.  If  the  percolation  factor  be  symbolized  by  c,  then  L,  or 
the  length  of  enforced  percolation,  will  equal  c  H,  II  being  the  head  of 
water.  The  factor  c  will  vary  in  value  with  the  quality  of  the  sand. 

Fig.  88  represents  a  case  similar  in  every  respect  to  the  last 
except,  instead  of  a  dam  of  earth,  the  obstruction  consists  of  a 
vertical  wall  termed  the  weir  or  drop  wall,  having  a  horizontal 
impervious  floor  attached  thereto,  which  appendage  is  necessary  to 
prevent  erosion  of  the  bed  by  the  current  of  falling  water. 

At  the  stage  of  maximum  pressure  the  head  water  will  be 
level  with  the  crest,  and  the  level  of  the  tail  water  that  of  the  floor; 
consequently  the  hydraulic  gradient  will  be  HB,  which  is  also  the 
piezometric  line  and  as  in  the  previous  case  of  the  pipe  line,  Fig. 


Fig.  88.     Diagram  Showing  Design  of  Profile  to  Reduce  Percolation 

86,  the  ordinates  of  the  triangle  HAB  will  represent  the  upward 
hydrostatic  pressure  on  the  base  of  the  weir  wall  and  of  the  floor. 

117.  Criterion  for  Safety  of  Structure.  The  safety  of  the 
structure  is  evidently  dependent  on  the  following  points: 

First,  the  weir  wall  must  be  dimensioned  to  resist  the  overturn- 
ing moment  of  the  horizontal  water  pressure.  This  has  been  dealt 
with  in  a  previous  section.  Second,  the  thickness,  i.  e.,  the  weight 
of  the  apron  or  floor  must  be  such  that  it  will  be  safe  from  being 
blown  up  or  fractured  by  the  hydrostatic  pressure;  third,  the  base 
length,  or  that  of  the  enforced  percolation  L,  must  not  be  less  than 
cH,  the  product  of  the  factor  c  with  the  head  H.  Fourth,  the 
length  of  the  masonry  apron  and  its  continuation  in  riprap  or  con- 
crete book  blocks  must  be  sufficient  to  prevent  erosion. 

It  is  evident  that  the  value-  of  this  factor  c,  must  vary  with  the 
nature  of  the  sand  substratum  in  accordance  with  its  qualities  of 


DAMS  AND  WEIRS  155 

fineness  or  coarseness.  Fine  light  sand  will  be  closer  in  texture, 
passing  less  water  under  a  given  head  than  a  coarser  variety,  but  at 
the  same  time  will  be  disintegrated  and  washed  out  under  less 
pressure.  Reliable  values  for  c,  on  which  the  design  mainly  depends, 
can  only  be  obtained  experimentally,  not  from  artificial  experiments, 
but  by  deduction  from  actual  examples  of  weirs;  among  which  the 
most  valuable  are  the  records  of  failures  due  to  insufficiency  in 
length  of  percolation.  From  these  statistics  a  safe  value  of  the 
relation  of  L  to  H,  the  factor  c,  which  is  also  the  sine  of  the 
hydraulic  gradient,  can  be  derived. 

118.  Adopted  Values  of  Percolation  Coefficient.  The  follow- 
ing values  of  c  have  been  adopted  for  the  specified  classes  of  sand. 

Class  I:  River  beds  of  light  silt  and  sand,  of  which  60  per  cent 
passes  a  100-mesh  sieve,  as  those  of  the  Nile  or  Mississippi;  percola- 
tion factor  c  =  18. 

Class  II:  Fine  micaceous  sand  of  which  80  per  cent  of  the 
grains  pass  a  75-mesh  sieve,  as  in  Himalayan  rivers  and  in  such  as 
the  Colorado;  c=  15. 

Class  III:  Coarse-grained  sands,  as  in  Central  and  South 
India  ;c  =  12. 

Class  IV:  Boulders  or  shingle  and  gravel  and  sand  mixed;  c 
varies  from  9  to  5. 

In  Fig.  88  if  the  sand  extended  only  up  to  the  level  C,  the  length 
of  percolation  would  be  CD,  the  rise  from  D  to  B  not  being  counted 
in.  In  that  case  the  area  of  hydrostatic  pressure  acting  beneath 
the  floor  would  be  the  triangle  II A  B.  As,  however,  a  layer  of  sand 
from  A  to  C  interposes,  the  length  will  be  ACD,  and  outline  II \B. 
The  step  H  HI  occurring  in  the  outline  is  due  to  the  neutralization 
of  head  symbolized  by  h,  effected  in  the  depth  AC.  Supposing 
A  C  to  be  6  feet  and  the  percolation  factor  to  be  12,  then  the  step 
in  the  pressure  area,  equal  to  h,  will  be  6  4- 12  =  6  inches.  The 
resulting  gradient  II iB  will,  however,  be  flatter  than  1  in  12;  conse- 
quently the  termination  of  the  apron  can  be  shifted  back  to  BiDi, 
IIiBi,  being  parallel  to  HE;  in  which  case  the  area  of  hydrostatic 
pressure  will  be  H\AEi.  The  pressure  at  any  point  on  the  base  is 
represented  by  the  ordinates  of  the  triangle  or  area  of  pressure. 
Thus  the  upward  pressure  at  E,  below  the  toe  of  the  drop  wall, 
where  the  horizontal  apron  commences,  is  represented  by  the  line 


156  DAMS  AND  WEIRS 

FG.  Supposing  the  head  HA  to  be  10  feet,  then  the  total  required 
length  of  percolation  will  be  cll  =  12X10  =  120  feet.  This  is  the 
length  A  CD  i .  The  neutralization  of  head,  h,  effected  by  the  enforced 
percolation  between  II  and  G  is  represented  by  GJ,  and  supposing 
the  base  width  of  the  drop  wall  CE  to  be  9  feet,  AC  being  6  feet, 

/*  =  6±9  =  1jL  feet>     The  upward  pressure   FG  is    (J/-fe)  =  10-li 
\.2i 

=  8|feet. 

The  stepped  upper  line  bounding  the  pressure  area  as  has  been 
noted  in  Part  I,  is  termed  the  piezometric  line,  as  it  represents  the 
level  to  which  water  would  rise  if  pipes  were  inserted  in  the  floor. 
It  is  evident  from  the  above  that  when  no  vertical  depressions 
occur  in  the  line  of  travel  that  the  piezometric  line  will  coincide 
with  the  hydraulic  gradient  or  virtual  slope ;  when,  however,  vertical 
depressions  exist,  reciprocal  steps  occur  in  the  piezometric  line, 
which  then  falls  below  the  hydraulic  grade  line.  The  piezometric 
line  is  naturally  always  parallel  to  the  latter.  The  commencement 
of  the  floor  at  E  is  always  a  critical  point  in  the  design  as  the  pres- 
sure is  greatest  here,  diminishing  to  zero  at  the  end. 

119.  Simplifying  the  Computations.  In  the  same  way  that 
the  water  pressure  is  represented  by  the  head  producing  it,  the 
common  factor  w,  or  the  unit  weight  of  water,  may  also  be  elimi- 
nated from  the  opposing  weight  of  the  floor.  The  weight  of  the 
masonry,  therefore,  is  represented  by  its  thickness  in  the  same  way 
as  the  pressure,  and  if  t  be  the  thickness  of  the  floor,  tp  will  represent 
its  weight.  Now  the  floor  lies  wholly  below  low  water  level.  Con- 
sequently, in  addition  to  the  external  hydrostatic  pressure  repre- 
sented by  //,  due  to  the  head  of  water  upheld,  there  is  the  buoyancy 
due  to  immersion.  The  actual  pressure  on  the  base  CDi  is  really 
measured  by  HC,  not  HA.  Thus  if  a  vertical  pipe  were  inserted 
in  the  floor  the  water  would  rise  up  to  the  piezometric  line  and  be 
in  depth  the  ordinate  of  the  pressure  area  plus  the  thickness  of  the 
floor.  But  it  is  convenient  to  keep  the  hydrostatic  external  pres- 
sure distinct  from  the  effect  of  immersion.  This  latter  can  be 
allowed  for  by  reduction  in  the  weight  of  these  parts  of  the  struc- 
ture that  lie  below  L.  W.  L.  See  sections  52  and  53,  Part  I. 

Effect  of  Immersion.  When  a  body  is  immersed  in  a  liquid  it 
loses  weight  to  the  extent  of  the  weight  of  the  liquid  displaced. 


DAMS  AND  WEIRS 


157 


Thus  the  unit  weight  of  a  solid  is  wp.    When  immersed,  the  unit 
weight  will  be  w (p  —  1).     As  w  is   a  discarded   factor,  the   unit 

weight  being  represented 
only  by  p,  the  specific 
gravity,  the  weight  of 
the  floor  in  question  will 
be  /(p  —  1)  if  immersed. 
We  have  seen  that  the 
hydrostatic  pressure  act- 
ing at  F  is  8 J  feet,  To 
meet  this  the  weight,  or 
effective  thickness,  of 
the  floor  must  be  equal 
to  8|  feet  of  water  +J 
|  for  safety,  or,  in  sym- 


Assuming  a  value  for  p 
of  2,  the  thickness  re- 
quired to  counterbal- 
ance the  hydrostatic 
pressure  will  be 

*  =  8fX-^  =  11.6  feet 

o 

The     formula     for 
thickness  will  then  stand : 


i-KfEi*) 


(33) 

S 

Uplift  on  Fore 
Apron.  It  is  evident 
that  in  Fig.  88  the  long 
floor  is  subjected  to  a 
very  considerable  uplift 
measured  by  the  area 
HAS,  the  weight  of  the 
apron  also  is  reduced  in 

the  ratio  of  p  :  (p  —  1)  as  it  lies  below  L.W.L.,  consequently  it  will 
have  to  be  made  as  already  noted  of  a  depth  of  11.6  feet  which  is  a 


158  DAMS  AND  WEIRS 

quite  impossible  figure.  The  remedy  is  either  to  make  the  floor  porous 
in  which  case  the  hydraulic  gradient  will  fall  below  1  in  12,  and  failure 
will  take  place  by  piping,  or  else  to  reduce  the  effective  head  by  the] 
insertion  of  a  rear  apron  or  a  vertical  curtain  wall  as  has  been  already 
mentioned  in  section  53,  Part  I.  In  these  submerged  weirs  on 
large  rivers  and  in  fact  in  most  overfall  dams  a  solid  fore  apron  is 
advisable.  The  length  of  this  should  however  be  limited  to  abso- 
lute requirements.  This  length  of  floor  is  a  matter  more  of  individual 
judgment  or  following  successful  precedent  than  one  of  precise 
estimation. 

The  following  empirical  rule  which  takes  into  account  the  nature 
of  the  sand  as  well  as  the  head  of  wrater  is  believed  to  be  a  good  guide 
in  determining  the  length  of  fore  apron  in  a  weir  of  this  type,  it  is 

L  =  3^I^H  (34) 

In  the  case  of  Fig.  89,  the  head  is  10  feet  and  c  is  assumed  at  12,  conse- 
quently, L  =  3Vl20  =  33  feet,  say  36,  or  3c.  In  Fig.  89  this  length 
of  floor  has  been  inserted.  Now  a  total  length  of  percolation  of  12 
times  the  head,  or  120  feet  =  10c  is  required  by  hypothesis,  of  this 
3c  is  used  up  by  the  floor  leaving  7c  to  be  provided  by  a  rear  apron 
and  curtain.  Supposing  the  curtain  is  made  a  depth  equal  to  IJc, 
this  will  dispose  of  3  out  of  the  7  (for  reasons  to  be  given  later), 
leaving  4  to  be  provided  for  by  the  rear  apron,  the  length  of  which, 
counting  from  the  toe  of  the  wreir  wrall,  is  made  4c  or  48  feet.  The 
hydraulic  gradient  starts  from  the  point  H'  which  is  vertically 
above  that  of  ingress  A.  At  the  location  of  the  vertical  diaphragm 
of  sheet  piling,  a  step  takes  place  owing  to  the  sudden  reduction  of 
head  of  3  feet,  the  obstruction  being  3c  in  length  counting  both 
sides.  From  here  on,  the  line  is  termed  the  piezometric  line  and  the 
pressure  area  is  the  space  enclosed  between  it  and  the  floor.  The 
actual  pressure  area  would  include  the  floor  itself,  but  this  has  been 
already  allowed  for  in  reduction  of  weight,  its  s.g.  being  taken  as 
unity  instead  of  2. 

The  uplift  on  the  weir  wall  is  the  area  enclosed  between  its 
base  and  the  piezometric  line,  In  calculating  overturning  moment, 
if  this  portion  were  considered  as  having  lost  weight  by  immersion 
it  would  not  quite  fully  represent  the  loss  of  effective  weight  due  to 
uplift,  because  above  the  floor  level  the  profile  of  the  weir  wall  is 


DAMS  AND  WEIRS  159 

not  rectangular,  while  that  of  the  pressure  area  is  more  nearly  so. 
The  foundation  could  be  treated  this  way,  the  superstructure  above 
AF  being  given  full  s.g.  and  the  uplift  treated  as  a  separate  vertical 
force  as  was  the  case  in  Fig.  40,  Part  I. 

120.  Vertical  Obstruction  to  Percolation.     Now  with  regard 
to  the  vertical  obstruction,  when  water  percolates  under  pressure 
beneath  an  impervious  platform  the  particles  are  impelled  upward 
by  the  hydrostatic  pressure  against  the  base  of  the  dam  and  also 
there  is  a  slow  horizontal  current  downstream.    The  line  of  least 
resistance  is  along  the  surface  of  any  solid  in  preference  to  a  shorter 
course  through  the  middle  of  the  sand,  consequently  when  a  vertical 
obstruction  as  a  curtain  wall  of  masonry  or  a  diaphragm  of  sheet 
piling  is  encountered  the  current  of  water  is  forced  downward  and 
the  obstruction  being  passed  it  ascends  the  other  side  up  to  the  base 
line  which  it  again  follows.    The  outer  particles  follow  the  lead 
of  the  inner  as  is  shown  by  the  arrows  in  Fig.  89.    The  value  of  a 
vertical  obstruction  is  accordingly  twice  that  of  a  similar  horizontal 
length  of  base.     Valuable  corroboration  of  the  reliability  of  the 
theory  of  percolation  adopted,  particularly  with  regard  to  reduction 
of  head  caused  by  vertical  obstruction,  has  been  received,  while  this 
article  was  on  the  press,  from  a  paper  in  the  proceeedings  of  the 
American  Society  of  Civil  Engineers  entitled  "The  Action  of  Water 
under  Dams"  by  J.  B.  T.  Coleman,  which  appeared  in  August,  1915. 
The  practical  value  of  the  experiments,  however,   is  somewhat 
vitiated  by  the  smallness  of  the  scale  of  operations  and  the  dispro- 
portion in  the  ratio  H:  L  to  actual  conditions.     The  length  of  base 
of  the  dam  experimented  on  should  not  be  less  than  50  feet  with  a 
head  of  5  feet. 

121.  Rear  Apron.     The  extension  of   the  floor  rearward   is 
termed  the  rear  apron.     Its  statical  condition  is  peculiar,  not  being 
subject  to  any  upward  hydrostatic  pressure  as  is  the  case  with  the 
fore  apron  or  floor.     Inspection  of  the  diagram,  Fig.  89,  will  show 
that  the  water  pressure  acting  below  the  floor  is  the  trapezoid  enclosed 
between  the  piezometric  line  and  the  floor  level;  \vhereas  the  down- 
ward pressure  is  represented  by  the  rectangle  HiA\HA,  which  is 
considerably  larger.     Theoretically  no  weight  is  required  in  the 
rear  apron,  the  only  proviso  being  that  it  must  be  impervious  and 
have  a  water-tight  connection  with  the  weir  wall,  otherwise  the 


160  DAMS  AND  WEIRS 

incidence  of  //  may  fall  between  the  rear  apron  and  the  rest  of  the 
work,  rendering  the  former  useless.  Such  a  case  has  actually 
occurred.  It  is,  however,  considered  that  the  rear  apron  must  be 
of  a  definite  weight,  as  otherwise  the  percolation  of  water  under- 
neath it  would  partake  of  the  nature  of  a  surface  flow,  and  so  pre- 
vent any  neutralization  of  head  caused  by  friction  in  its  passage 
through  sand.  Consequently,  the  effective  thickness,  or  rather 
t  (p  —  1)  should  not  be  less  than  four  feet.  Its  level  need  not  be  the 
same  as  that  of  the  fore  apron  or  floor.  In  fact,  in  some  cases  it 
has  been  constructed  level  or  nearly  so  with  the  permanent  crest 


•  «Tll  * 


Fig.  90.     View  of  Grand  Barrage  over  Nile  River 

of  the  drop  wall.  But  this  disposition  has  the  effect  of  reducing 
the  coefficient  of  discharge  over  the  weir  and  increasing  the  afflux 
or  head  water  level,  which  is  open  to  objection.  The  best  position 
is  undoubtedly  level  with  the  fore  apron. 

Another  point  in  favor  of  the  rear  apron  is  the  fact  that  it  is 
free  from  either  hydrostatic  pressure  or  the  dynamic  force  of  falling 
water,  to  which  the  fore  apron  is  subject;  it  can,  therefore,  be  con- 
structed of  more  inexpensive  material.  Clay  consolidated  when 
wet,  i.e.,  puddle,  is  just  as  effective  in  this  respect  as  the  richest 
cement  masonry  or  concrete,  provided  it  is  protected  from  scour 
where  necessary  by  an  overlay  of  paving  or  riprap,  and  has  a  reli- 


DAMS  AND  WEIRS 


161 


able  connection  with  the  drop  wall  and  the  rest  of  the  work.  In  old 
works  these  properties  of  the  rear  apron  were  not  understood,  and 
the  stanching  of  the  loose  stone  rear  apron  commonly  provided,  was 
left  to  be  effected  by  the  natural  deposit  of  silt.  This  deposit 
eventually  does  take  place  and  is  of  the  greatest  value  in  increasing 
the  statical  stability  of  the  weir,  but  the  process  takes  time,  and 
until  complete,  the  work  is  liable  to  excess  hydrostatic  pressure  and 
an  insufficient  length  of  enforced  percolation,  which  would  allow 
piping  to  take  place  and  the  foundation  to  be  gradually  undermined. 
122.  First  Demonstration  of  Rear  Apron.  Th^  value  of  an 
impervious  rear  apron  was  first  demonstrated  in  the  repairs  to  the 
Grand  Barrage  over  the  Nile,  Fig.  90,  some  time  in  the  eighties. 
This  old  work  was  useless  owing  to  the  great  leakage  that  took 
place  whenever  the  gates  were  lowered  and  a  head  of  water  applied. 
In  order  to  check  this  leakage,  instead  of  driving  sheet  piling,  which 


Fig.  91.     Section  Showing  Repairs  Made  on  the  Grand  Barrage 

it  was  feared  would  shake  the  foundations,  an  apron  of  cement 
masonry  240  feet  wide  and  3.28  feet  thick,  Fig.  91,  was  constructed 
over  the  old  floor,  extending  upstream  82  feet  beyond  it.  This 
proved  completely  successful.  By  means  of  pipes  set  in  holes 
drilled  in  the  piers,  cement  mortar  was  forced  under  pressure  into 
all  the  interstices  of  the  rubble  foundations,  filling  up  any  hollows 
that  existed,  thus  completely  stanching  the  foundations.  So  effec- 
tually wras  the  structure  repaired  that  it  was  rendered  capable  of 
holding  up  about  13  feet  of  Water;  whereas,  prior  to  reconstruction, 
it  was  unsafe  with  a  head  of  a  little  over  three  feet.  The  total 
length  of  apron  is  238  feet,  of  which  82  feet  projects  upstream 
beyond  the  original  floor  and  44  feet  downstream,  below  the  floor 
itself,  the  latter  having  a  width  of  112  feet.  The  head  H  being 

238 
13  feet,  and  L  being  238  feet,  c,  or  the  percolation  factor  is  ~^-  = 

io 

18,  which  is  the  exact  value  assigned  for  Nile  sand  in  Class  I, 


162  DAMS  AND  WEIRS 

section  118.     This  value  was  not  originally  derived  from  the  Grand 
Barrage,  but  from  another  work. 

The  utility  of  this  barrage  has  been  further  augmented  by  the 
construction  of  two  subsidiary  weirs  below  it,  see  Figs.  92  and  106, 
across  the  two  branches  of  the  Nile  delta.  These  are  ten  feet  high 
and  enable  an  additional  height  of  ten  feet  to  be  held  up  by 
the  gates  of  the  old  barrage,  the  total  height  being  now  22 J  feet. 
The  increased  rise  in  the  tail  water  exactly  compensates  for  the 
additional  head  on  the  wrork  as  regards  hydrostatic  pressure,  but 
the  moment  of  the  water  pressure  on  the  base  of  the  masonry  piers 


Fig.  92.     Plan  of  Grand  Barrage  over  Nile  River  Showing  Also  Location 
of  Damietta  and  Rosetta  Weirs 

will  be  largely  increased,  viz,  as  from  ]33  to  223  — 103,  or  from  2197 
to  9648. 

The  barrage,  which  is  another  word  for  "open  dam"  or  bulk- 
head dam,  is,  however,  of  very  solid  and  weighty  construction,  and 
after  the  complete  renewal  of  all  its  weak  points  is  now  capable  of 
safely  enduring  the  increased  stress  put  upon  the  superstructure. 
We  have  seen,  in  section  119,  that  the  width  of  the  impervious 
fore  apron  should  be  L  =  3^cII,  formula  (34).  This  width  of  the 
floor  is  affected  by  two  considerations,  first,  the  nature  of  the  river 
bed,  which  can  best  be  represented  by  its  percolation  factor  c  and 
second,  by  the  height  of  the  overfall  including  the  crest  shutters 
if  any,  which  will  be  designated  by  Ha  to  distinguish  it  from  H, 
which  represents  the  difference  between  head  and  tail  water  and  also 


DAMS  AND  WEIRS 


163 


TABLE  I 
Showing  Actual  and  Calculated  Values  of  Li  or  Talus  Width 

Formula  (35),  Lt 


RIVER 

NAME  OF  WORK 

TYPE 

C 

Hb 

Q 

LENGTH  LI 

Calculated 

Actual 

Ganges 

Narora 

A 

15 

10 

75 

150 

140-170 

Coleroon 

Coleroon    > 

A 

12 

4| 

100 

92 

72 

Vellar 

Pelandorai 

A 

9 

11 

100 

108 

101 

Tampra- 
parni 

Srivakantham 

A 

12 

6 

90 

102 

106 

Chenab 

Khanki 

B 

15 

7 

150 

182 

170 

Chenab 

Merala 

B 

15 

7 

150 

182 

203 

Jhelum 

Rasul 

B 

15 

6 

155 

160 

135 

Penner 

Adimapali 

B 

12 

8^ 

184 

172 

184 

Penner 

Nellore 

B 

12 

9 

300 

228 

232 

Penner 

Sangam 

B 

12 

10 

147 

168 

145 

Godaveri 

Dauleshwiram 

B 

12 

13 

100 

158 

217 

Jumna 

Okhla 

C 

15 

10 

140 

210 

210 

Kistna 

Beswada 

C 

12 

13 

223 

236 

220 

Son 

Dehri 

C 

12 

8 

66 

100 

96 

Mahanadi 

Jobra 

C 

12 

100 

140 

163 

143 

Madaya 

Madaya 

C 

12 

8 

280 

207 

235 

Colorado 

Laguna 

C 

15 

10 

below 
minimun 

140 

200 

Type  A  has  a  direct  overfall,  with  horizontal  floor  at  L.  W.  L.,  as  in 
Figs.  91,  93,  and  95. 

Type  B  has  breast  wall  followed  by  a  sloping  impervious  apron,  Figs. 
96  and  97. 

Type  C  has  breast  wall  followed  by  pervious  rock  fill,  with  sloping  surface 
and  vertical  body  walls,  Figs.  101  to  106. 

from  Hb  the  height  of  the  permanent  crest  above  L.  W.  L.  Tak- 
ing the  Narora  weir  a,*  standard,  a  length  of  floor  equal  to  3Vc# 
=  3Vl5Xl3  =  42  feet,  is  deemed  to  be  the  correct  safe  width  for 


164  DAMS  AND  WEIRS 

a  weir  13  feet  in  height;  where  the  height  is  more  or  less,  the 
width  should  be  increased  or  reduced  in  proportion  to  the  square 
root  of  the  height  and  that  of  the  factor  c. 

123.  Riprap  to  Protect  Apron.  Beyond  the  impervious  floor 
a  long  continuation  of  riprap  or  packed  stone  pitching  is  required. 
The  width  of  this  material  is  clearly  independent  of  that  appro- 
priate to  the  floor,  and  consequently  will  be  measured  from  the 
same  starting  point  as  the  floor,  viz,  from  the  toe  of  the  drop  wall. 
The  formula  for  overfall  weirs  is 

x^  (35) 

For  sloping  aprons,  type  B,  the  coefficient  of  c  will  be  11 
Then 


Ll=nc^W75  (35a) 

This  formula  is  founded  on  the  theory  that  the  distance  of  the 
toe  of  the  talus  from  the  overfall  will  vary  with  the  square  root  of 
the  height  of  the  obstruction  above  low  water,  designated  by  Hb) 
with  the  square  root  of  the  unit  flood  discharge  over  the  weir  crest 
q,  and  directly  with  c,  the  percolation  factor  of  the  river  sand.  The 
standard  being  these  values;  viz,  10,  75,  and  10,  respectively,  in 
Narora  weir.  This  height,  //&  is  equal  to  //  when  there  are  no 
crest  shutters,  and  is  always  the  depth  of  L.  W.  L.  below  the  per- 
manent masonry  crest  of  the  weir.  This  formula,  though  more  or 
less  empirical,  gives  results  remarkably  in  consonance  with  actual 
value,  and  will,  it  is  believed,  form  a  valuable  guide  to  design. 
Table  I  will  conclusively  prove  this.  As  nearly  all  the  weirs  of 
this  class  have  been  constructed  in  India,  works  in  that  country 
are  quoted  as  examples. 

124.  Example  of  Design  Type  A.  Another  example  of  design 
in  type  A  will  now  be  given,  Fig.  93,  the  dimensions  being  those  of 
an  actual  work,  viz,  the  Narora  weir  over  the  Ganges  River,  the 
design  being  thus  an  alternative  for  that  work,  the  existing  section 
of  which  is  shown  in  Fig.  95  and  discussed  in  section  125.  The 
data  on  which  the  design  is  based,  is  as  follows:  sand,  class  2;  per- 
colation factor  c  =  15;  //or  difference  between  head  and  low  water, 
the  latter  being  always  symbolized  by  L.  W.  L.,  13  feet,  unit  dis- 


DAMS  AND  WEIRS 


1G5 


charge  over  weir  q  =  75  second-feet,  the 
total  length  of  the  impervious  apron  and 
vertical  obstructions  will,  therefore,  have 
to  be  L  =  cII  =  15X13  =  195  feet. 

The  first  point  to  be  determined  is 
the  length  of  the  floor  or  fore'  apron. 
Having  fixed  this  length,  the  balance  of 
L  will  have  to  be  divided  among  the 
rear  apron  and  the  vertical  sheet  piling. 
It  is  essential  that  this  minimum  length 
be  not  exceeded,  as  it  is  clearly  of  advan- 
tage to  put  as  much  of  the  length  into 
the  rear  apron  as  possible,  owing  to  the 
inexpensive  nature  of  the  material  of 
which  it  can  be  constructed.  According 
to  formula  (2) ,  L  =  42  feet,  which  is  nearly 
equivalent  to  3c,  or  45  feet,  there  thus 
remains  lOc  to  be  proportioned  between 
the  rear  apron  and  the  vertical  curtain. 
If  the  latter  be  given  a  depth  of  2c,  or  30 
feet  the  length  of  travel  down  and  up  will 
absorb  4c,  leaving  6c,  or  90  feet  for  the 
rear  apron.  The  measurement  is  taken 
from  the  toe  of  the  drop  wall.  The 
neutralization  of  the  whole  head  of  13 
feet  is  thus  accomplished.  A  second  cur- 
tain will  generally  be  desirable  at  the 
extremity  of  the  fore  apron  as  a  pre- 
cautionary measure  to  form  a  protec- 
tion in  case  the  loose  riprap  downstream 
from  the  apron  is  washed  out  or  sinks. 
This  curtain  must  have  open  joints  to 
offer  as  little  obstruction  to  percolation 
as  possible.  The  outline  of  the  pressure 
area,  that  is  the  piezometric  line,  is 
drawn  as  follows:  cH=  195  feet  is  meas- 
ured horizontally  on  the  base  line  of  the 
pressure  area,  that  is,  at  L.  W.  L.  from  a 


DAMS  AND  WEIRS 


167 


§i| 

>  h 


5  i  * 

*>     !.2 

Q    2  -g 

Q  «« 


line  through  ^4  to  5.  The  point  7?  is 
then  joined  with  A  on  the  head  water 
level  at  the  commencement  of  the  rear 
apron.  The  hydraulic  gradient  will 
thus  be  1  in  15.  The  intersection  of 
this  line  BA  with  a  vertical  drawn 
through  the  first  line  of  curtain  is  the 
location  of  a  step  of  two  feet  equal  to 
the  head  absorbed  in  the  vertical  travel 
at  this  point.  Another  line  parallel  to 
the  hydraulic  gradient  is  now  drawn 
to  the  termination  of  the  fore  apron, 
this  completes  the  piezometric  line  or 
the  upper  outline  of  the  pressure  area. 
With  regard  to  the  floor  thickness 
at  the  toe  of  the  drop  wall,  the  value 
of  h,  or  loss  of  head  due  to  percola- 
tion under  the  rear  apron,  is  6  feet, 
from  the  rear  curtain,  4  feet;  total  10 
feet.  H-h  is,  therefore,  13-10  =  3. 
The  thickness  of  the  floor  according 
to  formula  (33)  where  (H-h)  =3  feet 
comes  to  f  X  3  =  4  feet,  the  value  of  p 
being  assumed  at  2.  The  floor  natur- 
ally tapers  toward  its  end  where  the 
uplift  is  nil.  The  thickness  at  this 
point  is  made  3  feet  which  is  about  the 
minimum  limit.  There  remains  now 
the  talus  of  riprap,  its  length  from 
formula  (35)  is 

L  =  10e\  ^\  ^  =  150  feet  =  lOc 

The  thickness  of  the  talus  is  generally  . 
four — often  five  feet — and  is  a  matter 
of  judgment  considering  the  nature  of 
the  material  used. 

125.     Discussion  of  Narora  Weir. 
The  Narora  weir  itself,  Fig.  94,  forms 


1G8  DAMS  AND  WEIRS 

a  most  instructive  object  lesson,  demonstrating  what  is  the  least 
correct  base  width,  or  length  of  percolation  consistent  with  absolute 
safety,  that  can  be  adopted  for  sands  of  class  2.  The  system  of 
analyzing  graphically  an  existing  work  with  regard  to  hydraulic 
gradient  is  exemplified  in  Fig.  95  under  three  separate  conditions; 
first,  as  the  work  originally  stood,  with  a  hydraulic  gradient  of  1  in 
11.8;  second,  at  the  time  of  failure,  when  the  floor  and  the  grouted 
riprap  blew  up.  On  this  occasion  owing  to  the  rear  apron  having 
been  washed  away  by  a  flood  the  hydraulic  grade  fell  to  1  in  8; 
third,  after  the  extension  of  the  rear  apron  and  curtailment  of  the 
fore  apron  had  been  effected.  Under  the  first  conditions  the  hori- 
zontal component  of  the  length  of  travel  or  percolation  L  from  A 
to  E  is  123  feet.  The  total  length  is  made  up  of  three  parts:  First, 
a  step  down  and  up  in  the  foundation  of  the  drop  wall  of  7  feet ; 
second,  a  drop  down  and  up  of  12  feet  either  side  of  the  downstream 
curtain  wall;  third,  the  horizontal  distance  123  as  above.  The  rise 
at  the  end  of  the  floor  is  neglected.  The  total  value  of  L  is  then 
123+7+12  +  12-154.  This  is  set  out  on  a  horizontal  line  to  the 
point  C.  AC  is  then  the  hydraulic  gradient. 

This  demonstrates  that  the  hydraulic  gradient  was  originally 
something  under  1  in  12,  and  in  addition  to  this  the  floor  is  very 
deficient  in  thickness.  The  hydrostatic  pressure  on  the  floor  at  the 
toe  of  the  drop  wall  is  8  feet.  To  meet  this  the  floor  has  a  value  of 
tp  of  only  5  feet.  The  specific  gravity  of  the  floor  will  not  exceed 
2,  as  it  was  mostly  formed  of  broken  brick  concrete  in  hydraulic 
mortar.  The  value  of  p  —  1  will,  therefore,  be  unity,  the  floor  being 
submerged.  In  spite  of  this,  the  work  stood  intact  to  all  external 
appearance  for  twenty  years,  when  a  heavy  freshet  in  the  river  set 
up  a  cross  current  which  washed  out  that  portion  of  the  rear  apron 
nearest  the  drop  wall,  thus  rendering  the  rest  useless,  the  connection 
having  been  severed. 

On  this  occurrence,  failure  at  once  took  place,  as  the  floor  had 
doubtless  been  on  the  point  of  yielding  for  some  time.  In  fact,  this 
state  of  affairs  had  been  suspected,  as  holes  bored  in  the  floor  very 
shortly  before  the  actual  catastrophe  took  place  showed  that  a  large 
space  existed  below  it,  full,  not  of  sand,  but  of  water.  Thus  the 
floor  was  actually  held  up  by  the  hydrostatic  pressure;  otherwise  it 
must  have  collapsed.  The  removal  of  the  rear  apron  caused  this 


DAMS  AND  WEIRS  169 

pressure  to  be  so  much  increased  that  the  whole  floor,  together  with 
the  grouted  pitching  below  the  curtain,  blew  up. 

The  hydraulic  gradient  BC  is  that  at  the  time  of  the  collapse. 
It  will  be  seen  that  it  is  now  reduced  to  1  in  8.  The  piezometric 
line  is  not  shown  on  the  diagram. 

In  restoring  the  work  the  rear  apron  was  extended  upstream  as 
shown  dotted  in  Fig.  95,  to  a  distance  of  80  feet  beyond  the  drop  wall, 
and  was  made  five  feet  thick.  It  was  composed  of  puddle  covered 
with  riprap  and  at  its  junction  with  the  drop  wall  was  provided 
with  a  solid  masonry  covering.  The  puddle  foundation  also  was 
sloped  down  to  the  level  of  the  floor  base  to  form  a  ground  connec- 
tion with  the  drop  wall.  At  its  upstream  termination  sheet  piling 
was  driven  to  a  depth  of  twelve  feet  below  floor  level. 

The  grouted  pitching  in  the  fore  apron  was  relaid  dry,  except 
for  the  first  ten  feet  which  was  rebuilt  in  mortar,  to  form  a  continua- 
tion of  the  impervious  floor.  Omitting  the  mortar  has  the  effect 
of  reducing  the  pressure  on  the  floor.  Even  then  the  uplift  would 
have  been  too  great,  so  a  water  cushion  2  feet  deep  was  formed  over 
the  floor  by  building  a  dwarf  wall  of  concrete  (shown  on  the  section) 
right  along  its  edge.  This  adds  1  foot  to  the  effective  value  of  tp. 
It  will  be  seen  that  the  hydraulic  gradient  now  works  out  to  1  in  15. 
A  value  for  c  of  15  has  been  adopted  for  similar  light  sands  from 
which  that  of  other  sands,  as  Classes  I  and  III,  have  been  deduced. 

It  will  be  noticed  that  the  crest  of  this  weir  is  furnished  with 
shutters  which  are  collapsible  when  overtopped  and  are  raised  by 
hand  or  by  a  traveling  crab  that  moves  along  the  crest,  raising  the 
shutters  as  it  proceeds.  The  shutters  are  3  feet  deep  and  some 
20  feet  long.  They  are  held  up  against  the  water  by  tension  rods 
hinged  to  the  weir,  and  at  about  J  the  height  of  the  shutter,  i.e., 
at  the  center  of  pressure. 

126.  Sloping  Apron  Weirs,  Type  B.  Another  type  of  weir, 
designated  B,  will  now  be  discussed,  in  which  there  is  no  direct 
vertical  drop,  the  fore  apron  not  being  horizontal  but  sloping  from 
the  crest  to  the  L.  W.  L.  or  to  a  little  above  it,  the  talus  beyond 
being  also  on  a  flat  slope  or  horizontal. 

In  the  modern  examples  of  this  type  which  will  be  examined, 
the  height  of  the  permanent  masonry  weir  wall  is  greatly  reduced, 
with  the  object  of  offering  as  little  obstruction  as  possible  to  the 


170 


DAMS  AND  WEIRS 


passage  of  flood  water.  The  canal  level  is  maintained  by  means  of 
deep  crest  shutters.  In  the  Khanki  weir,  Fig.  96,  the  weir  proper, 
or  rather  bar  wall,  is  7  feet  high  above  L.  W.  L.,  while  the  shutters 
are  6  feet  high.  It,  therefore,  holds  up  13  feet  of  water,  the  same 
as  was  the  case  with  the  Narora  weir. 

The  object  of  adopting  the  sloping  apron  is  to  avoid  construc- 
tion in  wet  foundations,  as  most  of  it  can  be  built  quite  in  the  dry 
above  L.  W.  L.  The  disadvantage  of  this  type  lies  in  the  con- 
striction of  the  waterway  below  the  breast  wall,  which  causes  the 
velocity  of  overfall  to  be  continued  well  past  the  crest.  With  a 
direct  overfall,  on  the  other  hand,  a  depth  of  7  feet  for  water  to 
churn  in  would  be  available  at  this  point.  This  would  check  the 
flow  and  the  increased  area  of  the  waterway  rendered  available 


Fig.  96. 


Profile  of  Khanki  Weir,  Showing  Restoration  Work  Similar  to  that  of 

Narora  Weir 


should  reduce  the  velocity.  For  this  reason,  although  the  action 
on  the  apron  is  possibly  less,  that  on  the  talus  and  river  bed  beyond 
must  be  greater  than  in  the  drop  wall  of  type  A. 

This  work,  like  the  former,  failed  for  want  of  sufficient  effective" 
base  length,  and  it  consequently  forms  a  valuable  object  lesson. 

As  originally  designed,  no  rear  apron  whatever,  excepting  a 
small  heap  of  stone  behind  the  breast  wall,  was  provided.  The 
value  of  L  up  to  the  termination  of  the  grouted  pitching  is  but  108 
feet;  whereas  it  should  have  been  cH  or  15X13  =  195  feet.  The 
.hydraulic  gradient,  as  shown  in  Fig.  96b,  is  only  1  in  8.3.  This 
neglects  the  small  vertical  component  at  the  breast  wall.  In  spite 
of  this  deficiency  in  effective  base  width,  the  floor,  owing  to  good 
workmanship,  did  not  give  way  for  some  years,  until  gradually 
increased  piping  beneath  the  base  caused  its  collapse. 


DAMS  AND  WEIRS  171 

Owing  to  the  raised  position  of  the  apron,  it  is  not  subject  to 
high  hydrostatic  pressure.  At  its  commencement  it  is  ten  feet 
below  the  summit  level  and  nine  feet  of  water  acts  at  this  point. 
This  is  met  by  four  feet  of  masonry  unsubmerged,  of  s.g.  2,  which 
almost  balances  it.  Thus  the  apron  did  not  blow  up,  as  was  the  case 
with  the  Narora  weir,  but  collapsed. 

Some  explanation  of  the  graphical  pressure  diagram  is  required, 
as  it  offers  some  peculiarities,  differing  from  the  last  examples. 
The  full  head,  or  II,  is  13  feet.  Owing,  however,  to  the  raised  and 
sloping  position  of  the  apron,  the  base  line  of  the  pressure  area  will 
not  be  horizontal  and  so  coincide  with  the  L.  W.  L.,  but  will  be  an 
inclined  line  from  the  commencement  a  to  the  point  b,  where  the 
sloping  base  coincides  with  the  L.  W.  L.  From  b  where  L.  W.  L.  is 
reached  onward,  the  base  will  be  horizontal.  With  a  sloping  apron 
the  pressure  is  nearly  uniform,  the  water-pressure  area  is  not  wedge 
shaped  but  approximates  to  a  rectangle.  The  apron,  therefore,  is 
also  properly  rectangular  in  profile,  whereas  in  the  overfall  type 
the  profile  is,  or  should  be,  that  of  a  truncated  wedge. 

127.  Restoration  of  Khanki  Weir.    After  the  failure  of  this 
work  the  restoration  was  on  very  similar  lines  to  that  of  the  Narora 
weir.     An  impervious  rear  apron,  seventy  feet  long,  was  constructed 
of  puddle  covered  with  concrete  slabs,  grouted  in  the  joints.     A  rear 
curtain  wall  consisting  of  a  line  of  rectangular  undersunk  blocks 
twenty  feet  deep,  wras  provided.    These  additions  have  the  effect 
of  reducing  the  gradient  to  1  in  16.     The  masonry  curtain  having 
regard  to  its  great  cost  is  of  doubtful  utility.     A  further  prolonga- 
tion of  the  rear  apron  or  else  a  line  of  sheet  piling  would,  it  is  deemed, 
have  been  equally  effective.     Reinforced-concrete    sheet  piling  is 
very  suitable  for  curtain  walls  in  sand  and  is  bound  to  supplant  the 
ponderous  and  expensive  block  curtain  walls  which  form  so  marked 
a  feature  in  Indian  works. 

128.  Merala  Weir.    Another  weir  on  the  same  principle  and 
quite  recently  constructed  is  the  Merala  weir  at  the  head  of  the 
same  historic  river,  the  Chenab,  known  as  the  "Hydaspes"  at  the 
time  of  Alexander  the  Great. 

This  weir,  a  section  of  which  is  given  in  Fig.  97,  is  located  in  the 
upper  reaches  of  the  river  and  is  subjected  to  very  violent  floods; 
consequently  its  construction  has  to  be  abnormally  strong  to  resist 


172 


DAMS  AND  WEIRS 


pt/oc\ 


* 


the  dynamic  action  of  the  water.  This 
is  entirely  a  matter  of  judgment  and 
no  definite  rules  can  possibly  be  given 
which  would  apply  to  different  condi- 
tions. From  a  hydrostatic  point  of 
view  the  two  lower  lines  of  curtain 
blocks  are  decidedly  detrimental  and 
could  well  be  cut  out.  If  this  were 
done  the  horizontal  length  of  travel 
or  percolation  will  come  to  140.  The 
head  is  12  or  13  feet.  If  the  latter,  c 
having  the  value  15  as  in  the  Khanki 
weir,  the  value  of  L  will  be  15  X 13  =  195 
feet.  The  horizontal  length  of  travel 
is  140  feet  and  the  wanting  55  feet 
will  be  just  made  up  by  the  rear  cur- 
tain. The  superfluity  of  the  two  fore 
lines  is  thus  apparent  with  regard  to 
hydrostatic  requirements.  The  long 
impervious  sloping  apron  is  a  necessity 
to  prevent  erosion. 

It  is  a  question  whether  a  line  of 
steel  interlocking  sheet  piling  is  equally 
efficient  as  a  curtain  formed  of  wells 
of  brickwork  12X8  feet  undersunk  and 
connected  with  piling  and  concrete 
filling.  The  latter  has  the  advantage 
of  solidity  and  weight  lacking  in  the 
former.  The  system  of  curtain  walls 
of  undersunk  blocks  is  peculiar  to 
India.  In  the  Hindia  Barrage,  in 
Mesopotamia,  Fig.  115,  interlocking 
sheet  piling  has  been  largely  employed 
in  places  where  well  foundations  would 
have  been  used  in  India.  This  change 
is  probably  due  to  the  want  of  skilled 
well  sinkers,  who  in  India  are  extremely 
expert  and  form  a  special  caste. 


DAMS  AND  WEIRS  173 

The  rear  apron,  in  the  Merala  weir  is  of  as  solid  construction 
as  the  fore  apron  and  is  built  on  a  slope  right  up  to  crest  level;  this 
arrangement  facilitates  discharge.  The  velocity  of  approach  must 
be  very  great  to  necessitate  huge  book  blocks  of  concrete  6X6X3 
feet  being  laid  behind  the  slope  and  beyond  that  a  40-foot  length 
of  riprap.  The  fore  apron  extends  for  93  feet  beyond  the  crest, 
twice  as  long  as  would  be  necessary  with  a  weir  of  type  A  under 
normal  conditions.  The  distance  L  of  the  talus  is  203  feet  against 
182  feet  calculated  from  formula  (35a).  That  of  the  lower  weir  at 
Khanki  is  170  feet.  This  shows  that  the  empirical  formula  gives 
a  fair  approximation. 

The  fore  apron  in  type  B  will  extend  to  the  toe  of  the  slope  or 
glacis.  It  is  quite  evident  that  the  erosive  action  on  a  sloping  apron 
of  type  B  is  far  greater  than  that  on  the  horizontal  floor  of  type  A;  the 

Summit  Level 


Probable  course  of  wafer  particles 


Fig.  98.     Diagram  Showing  Effect  of  Percolation  under  a  Wall 
Built   on   Sand 

uplift  however  is  less,  consequently  the  sloping  apron  can  be  made 
thinner  and  the  saving  thus  effected  put  into  additional  length. 
129.  Porous  Fore  Aprons0  The  next  type  of  weir  to  be  dealt 
with  is  type  C.  As  it  involves  some  fresh  points,  an  investigation 
of  it  and  the  principles  involved  will  be  necessary.  The  previous 
examples  of  types  A  and  B  have  been  cases  where  the  weir  has  as 
appendage  an  impervious  fore  apron  which  is  subject  to  hydro- 
static pressure.  There  is  another  very  common  type  which  will  be 
termed  C,  in  which  there  is  no  impervious  apron  and  the  material 
which  composes  the  body  of  the  weir  is  not  solid  masonry  but  a 
porous  mass  of  loose  stone  the  only  impervious  parts  being  narrow 
vertical  walls.  In  spite  of  this  apparent  contrariety  it  will  be 
found  that  the  same  principle,  viz,  that  of  length  of  enforced  per- 
colation, influences  the  design  in  this  type  as  in  the  others. 


174 


DAMS  AND  WEIRS 


Fig.  98  represents  a  wall  upholding  water  to  its  crest  and  resting 
on  a  pervious  substratum,  as  sand,  gravel,  or  boulders,  or  a  mixture 
of  all  three  materials.  The  hydraulic  gradient  is  AD;  the  upward 
pressure  area  ACD,  and  the  base  CD  is  the  travel  of  the  percolation. 
Unless  this  base  length  is  equal  to  AC  multiplied  by  the  percolation 
factor  obtained  by  experiment,  piping, will  set  in  and  the  wall  be 
undermined.  Now  as  shown  in  Fig.  99,  let  a  mass  of  loose  stone 
be  deposited  below  the  wall.  The  weight  of  this  stone  will  evi- 
dently have  an  appreciable  effect  in  checking  the  disintegration  and 
removal  of  the  sand  foundation.  The  water  will  not  have  a  free 
untrammeled  egress  at  D;  it  will,  on  the  contrary,  be  compelled  to 
rise  in  the  interstices  of  the  mass  to  a  certain  height  EE  determined 
by  the  extent  to  which  the  loose  stones  cause  obstruction  to  the  flow. 


Probable  course  of  water  particles  underneath  wall 
Fig.  99.     Effect  on  Percolation  Due  to  Stones  below  Weir  Wall  of  Fig.  98 

The  resulting  hydraulic  gradient  will  now  be  AE — flatter  than  AD, 
but  still  too  steep  for  permanency. 

In  Fig.  100,  the  wall  is  shown  backed  by  a  rear  apron  of  loose 
stone,  and  the  fore  apron  extended  to  F.  The  water  has  now  to 
filter  through  the  rear  apron  underneath  the  wall  and  up  through 
the  stone  filling  in  the  fore  apron.  During  this  process  a  certain 
amount  of  sand  will  be  washed  up  into  the  porous  body  and  the 
loose  stone  will  sink  until  the  combined  stone  and  sand  forms  a 
compact  mass,  offering  a  greater  obstruction  to  the  passage  of  the 
percolating  water  than  exists  in  the  sand  itself  and  possessing  far 
greater  resisting  power  to  disintegration.  This  will  cause  the  level 
of  water  at  E  to  rise  until  equilibrium  results.  When  this  is  the 
case  the  hydraulic  gradient  is  flattened  to  some  point  near  F.  If  a 
sufficiently  long  body  is  provided,  the  resulting  gradient  will  be 
equal  to  that  found  by  experiment  to  produce  permanent  equilibrium. 


DAMS  AND  WEIRS 


175 


The  mass  after  the  sinking  process  has  been 
finished  is  then  made  good  up  to  the  original  pro- 
file by  fresh  rock  filling.  At  F  near  the  toe  of 
the  slope  the  stone  offers  but  little  resistance 
either  by  its  weight  or  depth;  so  it  is  evident  that 
the  slope  of  the  prism  should  be  flatter  than  the 
hydraulic  gradient. 

The  same  action  takes  place  with  the  rear 
apron,  which  soon  becomes  so  filled  with  silt,  as 
to  be  impervious  or  nearly  so  to  the  passage  of 
water.  But  unless  silt  is  deposited  in  the  river 
bed  behind,  as  eventually  occurs  right  up  to  crest 
level,  the  thin  portion  of  the  rear  slope,  as  well  as 
the  similar  portion  of  the  fore  slope,  cannot  be 
counted  as  effective.  Consequently  out  of  the 
whole  base  length  this  part  GF,  roughly,  about 
one-quarter,  can  be  deemed  inefficient  as  regards 
length  of  enforced  percolation.  As  the  consoli- 
dated lower  part  of  the  body  of  the  weir  gains 
in  consistency,  it  can  well  be  subject  to  hydro- 
static pressure.  Consequently,  the  value  of  tp  of 
the  mass  should  be  in  excess  of  that  of  H—h,  just 
as  was  the  case  with  an  impervious  floor. 

130.  Porous  Fore  Aprons  Divided  by  Core 
Walls.  In  Fig.  101  a  still  further  development  is 
effected  by  the  introduction  of  vertical  body  or 
core  walls  of  masonry  in  the  pervious  mass  of  the 
fore  apron.  These  impervious  obstructions  mate- 
rially assist  the  stability  of  the  foundation,  so 
much  so  that  the  process  of  underscour  and  set- 
tlement which  must  precede  the  balancing  of  the 
opposing  forces  in  the  purely  loose  stone  mass 
need  not  occur  at  all,  or  to  nothing  like  the  same 
extent.  If  the  party  walls  are  properly  spaced, 
the  surface  slope  can  be  that  of  the  hydraulic 
gradient  itself  and  thus  ensure  equilibrium.  This 
is  clearly  illustrated  in  the  diagram.  The  water 
passing  underneath  the  wall  base  CD  will  rise  to 


DAMS  AND  WEIRS  177 

the  level  F,  the  point  E  being  somewhat  higher;  similar  percolation 
under  the  other  walls  in  the  substratum  will  fill  all  the  partitions  full 
of  water.  The  head  AC  will,  therefore,  be  split  up  into  four  steps. 

Value  of  Rear  Apron  Very  Great.  The  value  of  water  tightness 
in  the  rear  apron  is  so  marked  that  it  should  be  rendered  impervious 
by  a  thick  under  layer  of  clay,  and  not  left  entirely  to  more  or  less 
imperfect  surface  silt  stanching,  except  possibly  in  the  case  of  high 
dams  where  a  still  settling  pool  is  formed  in  rear  of  the  work. 

131.  Okhla  and  Madaya  Weirs.  In  Fig.  102  is  shown  a 
detailed  section  of  the  Okhla  rock-filled  weir  over  the  Jumna 
River,  India.  It  is  remarkable  as  being  the  first  rock-filled  weir 
not  provided  with  any  lines  of  curtain  walls  projecting  below 
the  base  line,  which  has  hitherto  invariably  been  adopted.  The 
stability  of  its  sand  foundation  is  consequently  entirely  depend- 
ent on  its  weight  and  its  effective  base  length.  As  will  be 
seen,  the  section  is  provided  with  two  body  walls  in  addition  to 
the  breast  wall.  The  slope  of  the  fore  apron  is  1  in  20.  It  is 
believed  that  a  slope  of  1  in  15  would  be  equally  effective,  a  hori- 
zontal talus  making  up  the  continuation,  as  has  been  done  in  the 
Madaya  weir,  Fig.  103,  which  is  a  similar  work  but  under  much 
greater  stress. 

The  head  of  the  water  in  the  Okhla  weir  is  13  feet,  with  shutters 
up  and  wreir  body  empty  of  water — a  condition  that  could  hardly 
exist.  This  would  require  an  effective  base  length,  L,  of  195  feet; 
the  actual  is  250  feet.  But,  as  noted  previously,  the  end  parts  of 
the  slopes  cannot  be  included  as  effective ;  consequently  the  hydraulic 
gradient  will  not  be  far  from  1  in  15.  The  weight  of  the  stone,  or 
tp,  exactly  balances  this  head  at  the  beginning,  as  it  is  10X1.3  =  13 
feet.  If  the  water  were  at  crest  level  and  the  weir  full  of  water, 
tp  would  equal  8  feet,  or  rather  a  trifle  less,  owing  to  the  lower  level 
of  the  crest  of  the  body  wall.  The  head  of  13  feet  is  broken  up 
into  four  steps.  The  first  is  3  feet  deep,  acting  on  a  part  of  the 
rear  apron  together  with  30  feet  of  the /ore  apron,  say,  1  in  15;  the 
rest  are  1  in  20.  A  slope  of  1  in  15  for  the  first  party  wall  would 
cut  the  base  at  a  point  40  feet  short  of  the  toe.  Theoretically  a 
fourth  party  wall  is  required  at  this  point,  but  practically  the  rip- 
rap below  the  third  dwarf  wall  is  so  stanched  with  sand  as  not  to 
afford  a  free  egress  for  the  percolation;  consequently  the  slope  may 


178 


DAMS  AND  WEIRS 


be  assumed  to  continue  on  to  its  inter- 
section with  the  horizontal  base.  As 
already  noted,  material  would  be  saved 
in  the  section  by  adopting  a  reliably 
stanch  rear  apron  and  reducing  the  fore 
slope  to  1  in  15,  with  a  horizontal  con- 
tinuation as  was  done  in  the  Madaya 
weir. 

Economy  of  Type  C.  This  type  C 
is  only  economical  where  stone  is  abun- 
dant. It  requires  little  labor  or  masonry 
work.  On  the  other  hand,  the  mass  of 
the  material  used  is  very  great,  much 
greater,  in  fact,  than  is  shown  by  the  sec- 
tion. This  is  owing  to  the  constant  sink- 
ing and  renewal  of  the  talus  which  goes 
on  for  many  years  after  the  first  construc- 
tion of  the  weir. 

The  action  of  the  flood  on  the  talus 
is  undoubtedly  accentuated  by  the  con- 
traction of  the  waterway  due  to  the  high 
sloping  apron.  The  fiood  velocity  20  feet 
below  the  crest  has  been  gaged  as  high 
as  18  feet  per  second.  This  would  be 
very  materially  reduced  if  the  A  type  of 
overfall  were  adopted,  as  the  area  of 
waterway  at  this  point  would  be  more 
than  doubled. 

132.  Dehri  Weir.  Another  typical 
example  of  this  class  is  the  Dehri  weir 
over  the  Son  River,  Fig.  104.  The  value 
of  L,  if  the  apexes  of  the  two  triangles  of 
stone  filling  are  deducted  and  the  .cur- 
tain walls  included,  comes  to  about  12  //, 
12  being  adopted  for  this  class  of  coarse 
sand .  The  curtain  walls,  each  over  1 2 , 500 
feet  long,  must  have  been  enormously 
costly.  From  the  experience  of  Okhla, 


DAMS  AND  WEIRS 

contemporary  work,  on  a  much  worse  class 
Is  of  sand,  curtain  protection  is  quite  unneces- 
d  isary  if  sufficient  horizontal   base   width   is 
r  j provided.    The  head  on  this  weir  is  10  feet, 
;  j  and  the  height  of  breast  wall  8  feet,  tp  is, 
therefore,   1.3X8  =  10.1,  which  is    sufficient, 
considering  that  the  full  head  will  not  act 
here.     The  lines  of  curtains  could  be  safely 
dispensed  with   if  the   following   alterations 
were  made:   (1)   Rear  apron  to  be  reliably 
stanched  in  order  to  throw  back  the  incidence 
of  pressure  and  increase  the  effective  base 
length;  (2)  three  more  body  walls  to  be  intro- 
duced; (3)  slope  1  in  12  retained,  but  base  to 
be  dredged  out  toward  apex  to  admit  of  no 
thickness    under    five  feet.     This  probably 
would  not  cause  any  increase  in  the  quantities 
of  masonry  above  what  they  now  are,  and 
would  entirely  obviate   the   construction    of 
nearly  five  miles  of  undersunk  curtain  blocks. 

133.  Laguna  Weir.    The  Laguna   weir 
over  the  Colorado  River,  the  only  example  of 
type  C  in  the  United  States  is  shown  in  Fig. 
105.     Compared  with  other  examples  it  might 
be  considered  as  somewhat  too  wide  if  regard 
is  had  to  its  low  unit  flood  discharge,  but  the 
inferior  quality  of  the   sand    of   this    river 
probably  renders  this  necessary.     The  body 
walls  are  undoubtedly  not  sufficiently  numer- 
ous to  be  properly  effective.    The  provision 
of  an  impervious  rear  apron  would  also  be 
advantageous. 

1 34.  Damietta  and  Rosetta  Weirs.    The 
location  of  the  Damietta  and  Rosetta  subsid- 
iary weirs,  Fig.  106,  which  have  been  rather 
recently  erected  below  the  old  Nile  barrage, 
is  shown  in  Fig.  92.     These  weirs  are  of  type 
C,  but  the  method  of  construction  is  quite 


179 


180 


DAMS  AND  WEIRS 


\82 


novel  and  it  is  this  alone  that  renders  this 
work  a  valuable  object  lesson.  The  deep 
foundation  of  the  breast  wall  was  built 
without  any  pumping,  all  material  having 
been  deposited  in  the  water  of  the  Nile 
River.  First  the  profile  of  the  base  was 
dredged  out,  as  shown  in  the  section.  Then 
the  core  wall  was  constructed  by  first 
depositing,  in  a  temporary  box  or  enclosure 
secured  by  a  few  piles,  loose  stone  from 
barges  floated  alongside.  The  whole  was 
then  grouted  with  cement  grout,  poured 
through  pipes  let  into  the  mass.  On  the 
completion  of  one  section  all  the  appliances 
were  moved  forward  and  another  section 
built,  and  so  on  until  the  whole  wall  was 
completed.  Clay  was  deposited  at  the  base 
of  the  core  wall  and  the  profile  then  made 
up  by  loose  stone  filling. 

This  novel  system  of  subaqueous  con- 
struction has  proved  so  satisfactory  that  in 
many  cases  it  is  bound  to  supersede  older 
methods.  Notwithstanding  these  innova- 
tions in  methods  of  rapid  construction,  the 
profile  of  the  weir  itself  is  open  to  the  objec- 
tion of  being  extravagantly  bulky  even  for 
the  type  adopted,  the  base  having  been 
dredged  out  so  deep  as  to  greatly  increase 
the  mean  depth  of  the  stone  filling. 

It  is  open  to  question  whether  a  row 
or  two  rows  of  concrete  sheet  piles  would 
not  have  been  just  as  efficacious  as  the  deep 
breast  wall,  and  would  certainly  have  been 
much  less  costly.  The  pure  cement  grout- 
ing was  naturally  expensive,  but  the  admix- 
ture of  sand  proved  unsatisfactory  as  the 
two  materials  of  different  specific  gravity 
separated  and  formed  layers;  consequently, 


DAMS  AND  WEIRS  181 

pure  cement  had  to  be  used.  It  may  be  noted  that  the  value  of  L 
here  is  much  less  than  would  be  expected.  At  Narora  weir  it  is  lie, 
or  165  feet.  Here,  with  a  value  of  c  of  18,  it  is  but  8Jc,  or  150  feet, 
instead  of  200  feet,  according  to  the  formula.  This  is  due  to  the 
low  flood  velocity  of  the  Nile  River  compared  with  the  Ganges. 

135.  The  Paradox  of  a  Pervious  Dam.  From  the  conditions 
prevailing  in  type  C  it  is  clear  that  an  impervious  apron  as  used  in 
types  A  and  B  is  not  absolutely  essential  in  order  to  secure  a  safe 
length  of  travel  for  the  percolating  subcurrent.  If  the  water  is 
free  to  rise  through  the  riprap  and  at  the  same  time  the  sand  in  the 
river  bed  is  prevented  from  rising  with  it,  the  practical  effect  is  the 
same  as  with  an  impervious  apron.  "Fount  aining",  as  spouting 
sand  is  technically  termed,  is  prevented  and  consequently  also 
"piping".  This  latter  term  defines  the  gradual  removal  of  sand 
from  beneath  a  foundation  by  the  action  of  the  percolating  under 
current.  Thus  the  apparent  paradox  that  a  length  of  filter  bed, 
although  pervious,  is  as  effective  as  a  masonry  apron  would  be. 
The  hydraulic  gradient  in  such  case  will  be  steeper  than  allowable 
under  the  latter  circumstance.  Filter  beds  are  usually  composed 
of  a  thick  layer  of  gravel  and  stone  laid  on  the  sand  of  the  river 
bed,  the  small  stuff  at  the  bottom  and  the  larger  material  at  the  top. 
The  ideal  type  of  filter  is  one  composed  of  stone  arranged  in  sizes  as 
above  stated  of  a  depth  of  4  or  5  feet  covered  with  heavy  slabs  or 
book  blocks  of  concrete;  these  are  set  with  narrow  open  intervals 
between  blocks  as  shown  in  Figs.  96  and  97.  Protection  is  thus 
afforded  not  only  against  scour  from  above  but  also  from  uplift 
underneath.  Although  the  subcurrent  of  water  can  escape  through 
a  filter  its  free  exit  is  hindered,  consequently  some  hydrostatic 
pressure  must  still  exist  below  the  base,  how  much  it  is  a  difficult 
matter  to  determine,  and  it  will  therefore  be  left  out  of  considera- 
tion. If  the  filter  bed  is  properly  constructed  its  length  should  be 
included  in  that  of  L  or  the  length  of  travel.  Ordinary  riprap, 
unless  exceptionally  deep,  is  not  of  much,  if  any,  value  in  this 
respect.  The  Hindia  Barrage  in  Mesopotamia,  Fig.  115,  section  145,  is 
provided  with  a  filter  bed  consisting  of  a  thick  layer  of  stone  65.5  feet 
wide  which  occurs  in  the  middle  of  the  floor.  The  object  of  this 
is  to  allow  the  escape  of  the  subcurrent  and  reduce  the  uplift  on 
the  dam  and  on  that  part  of  the  floor  which  is  impervious. 


182  DAMS  AND  WEIRS 

136.  Crest  Shutters.    Nearly  all  submerged  river  weirs  are 
provided  with  crest  shutters  3  to  6  feet  deep,  6  feet  being  the 
height  adopted  in  the  more   recent  works.     These  are  generally 
raised  by  means  of  a  traveling  crane  running  on  rails  just  behind 
the  hinge  of  the  gate.     When  the  shutters  are  tripped  they  fall 
over  this  railway.     In  the  case  of  the  Merala  weir,  Fig.  97,  the 
raising  of  the  shutters  is  effected  from  a  trolley  running  on  overhead 
wires  strung  over  steel  towers  erected  on  each  pier.     These  piers  or 
groins  are  500  feet  apart.     The  6-foot  shutters  are  3  feet  wide, 
held  up  by  hinged  struts  which  catch  on  to  a  bolt  and  are  easily 
released  by  hand  or  by  chains  worked  from  the  piers.     On  the 
Betwa  weir  the  shutters,  also  6  feet  deep,  are  automatic  in  action, 
being  hinged  to  a  tension  rod  at  about  the  center  of  pressure,  con- 
sequently when  overtopped  they  turn  over  and  fall.     Not  all  are 
hinged  at  the  same  height;  they  should  not  fall  simultaneously  but 
ease  the  flood  gradually.     The  advantage  of  deep  shutters  is  very 
great  as  the  permanent  weir  can  be  built  much  lower  than  other- 
wise would  be  necessary,  and  thus  offer  much  less  obstruction  to 
the  flood.     The  only  drawback  is  that  crest  shutters  require  a  resi- 
dent staff  of  experienced  men  to  deal  with  them. 

The  Laguna  weir,  Fig.  105,  has  no  shutters.  The  unit  flood 
discharge  of  the  Colorado  is,  however,  small  compared  with  that  of 
the  Indian  rivers,  being  only  22  second-feet,  whereas  the  Merala 
weir  discharges  150  second-feet  per  foot  run  of  weir,  consequently 
shutters  in  the  former  case  are  unnecessary. 

OPEN  DAMS  OR  BARRAGES 

137.  Barrage  Defined.    The    term  "open  dam",  or  barrage, 
generally  designates  what  is  in  fact  a  regulating  bridge  built  across 
a  river  channel,  and  furnished  with  gates  which  close  the  spans  as 
required.    They   are    partial    regulators,    the    closure    being   only 
effected  during  low  water.     When  the  river  is  in  flood,  the  gates 
are  opened  and  free  passage  is  afforded  for  flood  water  to  pass,  the 
floor  being  level  with  the  river  bed.     Weir  scouring  sluices,  which 
are  indispensable  adjuncts  to  weirs  built  over  sandy  rivers,  belong 
practically  to  the  same  category  as  open  dams,  as  they  are  also 
partial  regulators,  the  difference  being  that  they  span  only  a  por- 
tion of  the  river  instead  of  the  whole,  and  further  are  subject  to  great 


DAMS  AND  WEIRS  183 

scouring  action  from  the  fact  that  when  the  river  water  is  artificially 
raised  above  its  normal  level  by  the  weir,  the  downstream  channel 
is  empty  or  nearly  so. 

Function  of  Weir  Sluices.  The  function  of  weir  sluices  is  two- 
fold: First,  to  train  the  deep  channel  of  the  river,  the  natural 
course  of  which  is  obliterated  by  the  weir,  past  the  canal  head,  and 
to  retain  it  in  this  position.  Otherwise,  in  a  wide  river  the  low 
water  channel  might  take  a  course  parallel  to  the  weir  crest  itself, 
or  else  one  distant  from  the  canal  head,  and  thus  cause  the  approach 
channel  to  become  blocked  with  deposit. 

Second,  by  manipulating  the  sluice  gates,  silt  is  allowed  to 
deposit  in  the  slack  water  in  the  deep  channel.  The  canal  is  thereby 
preserved  from  silting  up,  and  when  the  accumulation  becomes 
excessive,  it  can  be  scoured  out  by  opening  the  gates. 

The  sill  of  the  weir  sluice  is  placed  as  low  as  can  conveniently 
be  managed,  being  generally  either  at  L.  W.  L.  itself,  or  somewhat 
higher,  its  level  generally  corresponding  with  the  base  of  the  drop 
or  breast  wall.  Thus  the  maximum  statical  head  to  which  the  work 
is  subjected  is  the  height  of  the  weir  crest  plus  that  of  the  weir 
shutters,  or  HI. 

The  ventage  provided  is  regulated  by  the  low-water  discharge 
of  the  river,  and  should  be  capable  of  taking  more  than  the  average 
dry  season  discharge.  In  one  case,  that  of  the  Laguna  weir,  where 
the  river  low  supply  is  deficient,  the  weir  sluices  are  designed  to  take 
the  whole  ordinary  discharge  of  the  river  excepting  the  highest 
floods.  This  is  with  the  object  of  maintaining  a  wide,  deep  channel 
which  may  be  drawn  upon  as  a  reservoir.  This  case  is,  however, 
exceptional. 

As  the  object  of  a  weir  sluice  is  to  pass  water  at  a  high  velocity 
in  order  to  scour  out  deposit  for  some  distance  to  the  rear  of  the 
work,  it  is  evident  that  the  openings  should  be  wide,  with  as  few 
obstructions  as  possible  in  the  way  of  piers,  and  should  be  open  at 
the  surface,  the  arches  and  platform  being  built  clear  of  the  flood 
level.  Further,  in  order  to  take  full  advantage  of  the  scouring 
power  of  the  current,  which  is  at  a  maximum  at  the  sluice  itself, 
diminishing  in  velocity  with  the  distance  to  the  rear  of  the  work, 
it  is  absolutely  necessary  not  only  to  place  the  canal  head  as  close 
as  possible  to  the  weir  sluices,  but  to  recess  the  head  as  little 


184 


DAMS  AND  WEIRS 


Scale  of  feet 

0       ZO      4O      6O     BO     100 

E'le^ahons  refer  h  Sea  Level 


Fig.  107.     Plan  of  Laguna  Weir-Scouring  Sluices 


DAMS  AND  WEIRS 


185 


as  practicable  behind  the  face  line  of  the  abutment  of  the  end 
sluice  vent. 

With  regard  to  canal  head  regulators  or  intakes,  the  regulation 
effected  by  these  is  entire,  not  partial,  so  that  these  works  are  sub- 
jected to  a  much  greater  statical  stress  than  weir  sluices,  and  conse- 
quently, for  convenience  of  manipulation,  are  usually  designed  with 
narrower  openings  than  are  necessary  or  desirable  in  the  latter. 
The  design  of  these  works  is,  however,  outside  the  scope  of  the 
subject  in  hand. 

138.  Example  of  Weir  Scouring  Sluice.  Fig.  107  is  an  excel- 
lent example  of  a  weir  scouring  sluice,  that  attached  to  the  Laguna 


Fig.  108. 


View  of  Yuma  Canal  and  Sluiceway  Showing  Sluice 
Gates  under   Construction 


weir,  the  profile  of  which  was  given  in  Fig.  105.  The  Yuma 
canal  intake  is  placed  clear  of  the  sandy  bed  of  the  river  on  a  rock 
foundation  and  the  sluiceway  in  front  of  it  is  also  cut  through  solid 
rock  independent  of  the  weir.  At  the  end  of  this  sluiceway  and  just 
past  the  intake  the  weir  sluices  are  located,  consisting  of  three  spans 
of  33|  feet  closed  by  steel  counterweighted  roller  gates  which  can  be 
hoisted  clear  of  the  flood  by  electrically  operated  winches.  The 
gates  are  clearly  shown  in  Fig.  108,  which  is  from  a  photograph 
taken  during  the  progress  of  the  wrork.  The  bed  of  the  sluiceway 
is  at  EL  138.0,  that  of  the  canal  intake  sill  is  147.0,  and  that  of  the 


186 


DAMS  AND  WEIRS 


weir  crest  151.0 — hence  the  whole  sluiceway  can  be  allowed  to  fill 
up  with  deposit  to  a  depth  of  9  feet,  without  interfering  with  the 


I  ,^3"ata.  weef/yfto<es x    , 


Fig.  109.     Plan  of  Weir  Sluices  for  Corbett  Dam  on  Shoshone  River,  Wyoming 

discharge  of  the  canal,  or  if  the  flashboards  of  the  intake  are  lowered 
the  sluiceway  can  be  filled  up  to  EL  156  which  is  the  level  of  the  top 
of  the  draw  gates,  i.e.,  18  feet  deep.  The  difference  between  high 


DAMS  AND  WEIRS 


187 


188 


DAMS  AND  WEIRS 


DAMS  AND  WEIRS 


189 


i      d 


water  above  and  below  the  sluice  gates 
is  1 1  feet,  consequently  when  the  gates 
are  lifted  immense  scour  must  take  place 
and  any  deposit  be  rapidly  removed. 
The  sluiceway  is  in  fact  a  large  silt 
trap. 

139.  Weir  Sluices  of  Corbett  Dam. 
The  weir  sluices  of  the  Corbett  dam  on 
the  Shoshone  River,  Wyoming,  are  given 
in  Figs.  109,  110,  and  111. 

The  canal  takes  out  through  a  tun- 
nel, the  head  of  which  has  necessarily  to 
be  recessed  far  behind  the  location  of  the 
weir  sluices.  Unless  special  measures 
were  adopted,  the  space  between  the 
sluice  gates  and  the  tunnel  head  would 
fill  up  with  sand  and  deposit  and  block 
the  entrance. 

To  obviate  this  a  wall  8  feet  high  is 
built  encircling  the  entrance.  A  "divide" 
wall  is  also  run  out  upstream  of  the  weir 
sluices,  cutting  them  off  from  the  weir 
and  its  approaches.  The  space  between 
these  two  walls  forms  a  sluiceway  which 
draws  the  current  of  the  river  in  a  low 
stage  past  the  canal  head  and  further 
forms  a  large  silt  trap  which  can  be 
scoured  out  when  convenient.  Only  a 
thin  film  of  surface  water  can  overflow 
the  long  encircling  wall,  then  it  runs  down 
a  paved  warped  slope  which  leads  it  into 
the  head  gates,  the  heavy  silt  in  suspen- 
sion being  deposited  in  the  sluiceway. 
This  arrangement  is  admirable. 

The  fault  of  the  weir  sluices  as  built 
is  the  narrowness  of  the  openings  which 
consist  of  three  spans  of  5  feet.  One 
span  of  12  feet  would  be  much  more 


190 


DAMS  AND  WEIRS 


effective.  In  modern  Indian  practice,  weir  sluices  on  large  rivers 
are  built  with  20  to  40  feet  openings. 

140.  Weir  Scouring  Sluices  on  Sand.  Weir  scouring  sluices 
built  on  pure  sand  on  as  large  rivers  as  are  met  with  in  India  are 
very  formidable  works,  provided  with  long  aprons  and  deep  lines 
of  curtain  blocks.  An  example  is  given  in  Fig.  112  of  the  so-termed 
undersluices  of  the  Khanki  weir  over  the 'Chenab  River  in  the  Pun- 
jab. The  spans  are  20  feet,  each  closed  by  3  draw  gates,  running 
in  parallel  grooves,  fitted  with  antifriction  wheels  (not  rollers),  lifted 
by  means  of  traveling  power  winches  which  straddle  the  openings 
in  which  the  grooves  and  gates  are  located. 

The  Merala  weir  sluices  of  the  Upper  Chenab  canal  have  8 
spans  of  31  feet,  piers  5J  feet  thick,  double  draw  gates  14  feet  high. 


Fig.  113.     View  of  Merala  Weir  Sluices,  Upper  Chenab  Canal 

These  are  lifted  clear  of  the  flood,  which  is  21  feet  above  floor,  by 
means  of  steel  towers  20  feet  high  erected  on  each  pier.  These  carry 
the  lifting  apparatus  and  heavy  counterweights.  These  gates,  like 
those  at  Laguna  weir,  Fig.  108,  bear  against  Stoney  roller  frames. 

Fig.  113  is  from  a  photograph  of  the  Merala  weir  sluices.  The 
work  is  a  partial  regulator,  in  that  complete  closure  at  high  flood  is 
not  attempted.  The  Upper  Chenab  canal  is  the  largest  in  the  world 
with  the  sole  exception  of  the  Ibramiyah  canal  in  JEgypt,  its  dis- 
charge being  12,000  second-feet.  Its  depth  is  13  feet.  The  capacity 
of  the  Ibramiyah  was  20,000  second-feet  prior  to  head  regulation. 

141.  Heavy  Construction  a  Necessity.  In  works  of  this 
description  solid  construction  is  a  necessity.  Light  reinforced  con- 
crete construction  would  not  answer,  as  weight  is  required,  not  only 


DAMS  AND  WEIRS 


191 


to  withstand  the  hydrostatic  pressure  but  the  dynamic  effects  of 
flood  water  in  violent  motion.     Besides  which  widely  distributed 


weight  is  undoubtedly  necessary  for  works  built  on  the  shifting 
sand  of  a  river  bed,  although  this  is  a  matter  for  which  no  definite 
rules  can  be  formed. 


192  DAMS  AND  WEIRS 

The  weir  sluices  at  Laguna  and  also  at  the  Corbett  dam,  are 
solid  concrete  structures  without  reinforcement. 

In  the  East,  generally,  reinforced  concrete  is  not  employed  nor 
is  even  cement  concrete  except  in  wet  foundations,  the  reason 
being  that  cement,  steel,  and  wood  for  forms  are  very  expensive 
items  whereas  excellent  natural  hydraulic  lime  is  generally  avail- 
able, skilled  and  unskilled  labor  is  also  abundant.  A  skilled  mason's 
wages  are  about  10  to  16  cents  and  a  laborer's  6  to  8  cents  for  a 
12-hour  day.  Under  such  circumstances  the  employment  of  rein- 
forced cement  concrete  is  entirely  confined  to  siphons  where  tension 
has  to  be  taken  care  of. 

In  America,  on  the  other  hand,  the  labor  conditions  are  such 
that  reinforced  concrete  which  requires  only  unskilled  labor  and  is 
mostly  made  up  by  machinery,  is  by  far  the  most  suitable  form  of 
construction  from  point  of  view  of  cost  as  well  as  convenience. 

This  accounts  for  the  very  different  appearance  of  irrigation 
works  in  the  East  from  those  in  the  West.  Both  are  suitable  under 
the  different  conditions  that  severally  exist. 

142.  Large  Open  Dams  across  Riversr  Of  open  dams  built 
across  rivers,  several  specimens  on  a  large  scale  exist  in  Egypt. 
These  works,  like  weir  sluices  are  partial  regulators  and  allow  free 
passage  to  flood  water. 

Assiut  Barrage.  In  the  Assiut  barrage,  Figs.  114  and  115,* 
constructed  across  the  Nile  above  the  Ibramiyah  canal  head 
in  lower  Egypt,  the  foundations  are  of  sand  and  silt  of  a  worse 
quality  than  is  met  with  in  the  great  Himalayan  rivers.  •_  The 
value  of  c  adopted  for  the  Nile  is  18,  against  15  for  Himalayan  rivers. 
This  dam  holds  up  5  meters  of  water,  the  head  or  difference  of 
levels  being  3  meters.  Having  regard  to  uplift,  the  head  is  the 
difference  of  levels  but  when  considering  overturning  moment,  on 

the  piers,  — —  is  the  moment,  H  and  h  being  the  respective 

o       6 

depths  of  water  above  and  below  the  gates.  It  is  believed  that  in 
the  estimation  of  the  length  of  travel  the  vertical  sheet  piling  was 
left  out  of  consideration.  Inspection  of  the  section  in  Fig.  115 

*  In  Figs.  115  and  116  and  in  the  discussion  of  these  problems  in  the  text,  the  metric  dimen- 
sions used  in  the  plans  of  the  works  have  been  retained.  Meters  multiplied  by  the  factor 
3.28  will  give  the  proper  values  in  feet. 


DAMS  AND  WEIRS 


193 


shows  that  the  foundation  is 
mass  cement  concrete,  10  feet 
deep,  on  which  platform  the 
superstructure  is  built.  This 
latter  consists  of  122  spans  of  5 
meters,  or  16  feet,  with  piers  2 
meters  thick,  every  ninth  being 
an  abutment  pier  4  meters  thick 
and  longer  than  the  rest.  This 
is  a  work  of  excessive  solidity 
the  ratio  of  thickness  of  piers 
to  the  span  being  .48,  a  propor- 
tion of  .33  S  would,  it  is  con- 
sidered be  better.  This  could 
be  had  by  increasing  the  spans 
to  6  meters,  or  20  feet  right 
through,  retaining  the  pier  thick- 
ness as  it  is  at  present. 

143.  General  Features  of 
River  Regulators.  All  these  river 
regulators  are  built  on  the  same 
general  lines,  viz,  mass  founda- 
tions of  a  great  depth,  an  arched 
highway  bridge,  with  spring  of 
arch  at  flood  level,  then  a  gap 
left  for  insertion  of  the  double 
grooves  and  gates,  succeeded  by 
a  narrow  strip  of  arch  sufficient 
to  carry  one  of  the  rails  of  the 
traveling  winch,  the  other  rest- 
ing on  the  one  parapet  of  the 
bridge. 

The  piers  are  given  a  bat- 
ter downstream  in  order  to  bet- 
ter distribute  the  pressure  on 
the  foundation.  The  resultant 
of  the  weight  of  one  span  com- 
bined with  the  horizontal  water 


194  DAMS  AND  WEIRS 

pressure  must  fall  within  the  middle  third  of  the  base  of  the  pier, 
the  length  of  which  can  be  manipulated  to  bring  this  about.  In 
this  case  it  does  so  even  with  increase  of  the  span  to  6  meters. 
This  combined  work  is  of  value  considered  from  a  military  point 
of  view,  as  affording  a  crossing  of  the  Nile  River;  consequently 
the  extreme  solidity  of  its  construction  was  probably  considered  a 
necessity. 

In  some  regulators  girders  are  substituted  for  arches,  in  others 
as  we  have  noted  with  regard  to  the  Merala  weir  sluices,  the  super- 
structure above  the  flood  line  is  open  steel  work  of  considerable 
height. 

144.  Stability  of  Assiut  Barrage.    The  hydraulic  gradient  in 
Fig.  115,  neglecting  the  vertical  sheet  piling,  is  drawn  on  the  profile 
and  is  the  line  AB,  the  horizontal  distance  is  43  meters  while  the 
head  is  3  meters.     The  slope  is  therefore  1  in  14  J.     The  uplift  is 
the  area  enclosed  between  AB  and  a  horizontal  through  B  which  is 
only  1.4  meters  at  its  deepest  part  near  the  gates.     Upstream  of 
the  gates  the  uplift  is  more  than  balanced  by  the  weight  of  water 
overlying  the  floor.     The  horizontal  travel  of  the  percolation  is  from 
A  to  B  plus  the  length  of  the  filter  as  explained  in  section    135. 

The  horizontal  travel  is  therefore  51J  meters  and  the  ratio— , or 
c,  is— ^-  =  17.2.  The  piezometric  line  has  also  been  shown,  includ- 

o 

ing  in  this  case  the  two  vertical  obstructions.  Their  effect  on  the 
uplift  is  very  slight,  owing  to  the  fore  curtain  which  raises  the  grade 
line.  The  slope  in  this  case  is  obtained  by  adding  the  vertical  to 
the  horizontal  travel,  i.e.,  from  B  to  D,  BC  and  CD  being  8  meters 
each  in  length,  AD  is  then  the  hydraulic  gradient  which  is  1  in  23. 
Steps  occur  at  points  b  and  c;  for  instance  the  line  AB  is  part  of 
AD,  the  line  be  is  parallel  to  AD  drawn  up  from  C,  and  the  line 
cB  is  similarly  drawn  up  from  B  forming  the  end  step. 

This  work  is  the  first  to  be  built  with  a  filter  downstream, 
which  has  the  practical  effect  of  adding  to  the  length  of  percolation 
travel  irrespective  of  the  hydraulic  grade. 

145.  The  Hindia  Barrage.    The  Hindia  barrage,  quite  recently 
erected  over  the  Euphrates  River  near  Bagdad,  is  given  in  Fig.  116. 
This  work,  which  was  designed  by  Sir  William  Willocks,  bears  a 


DAMS  AND  WEIRS 


195 


close  resemblance  to 
the  Egyptian  regu- 
lators, viz,  the  Assiut, 
the  Zifta,  and  other 
works  constructed 
across  the  river  Nile. 
The  piers  are  reduced 
to  1.50  meters  from 
the  2-meters  thick- 
ness in  the  Assiut 
dam,  Fig.  115,  and 
there  are  no  abut- 
ment piers,  conse- 
quently the  elevation 
presents  a  much 
lighter  appearance. 
The  ratio  of  thick- 
ness to  span  is  .3. 
In  order  to  reduce 
the  head  on  the  work, 
a  filter  bed  20  meters 
wide  is  introduced 
just  beyond  the  plat- 
form of  the  founda- 
tion of  the  regulat- 
ing bridge.  The 
upward  pressure  is 
thus  presumed  to  be 
nil  at  the  point  D. 
The  head  is  the  dis- 
tance between  the 
summer  supply  level 
upst^am,  and  that 
downstream  above 
the  subsidiary  weir, 
this  amounts  to  3.50 
meters.  The  length 
of  compulsory  travel 


196  DAMS  AND  WEIRS 

from  A  to  B  including  .50  meter  due  to  the  sheet  piling  is  36.50 

36  5 

meters.     AB  is  then  the  hydraulic  gradient,  which  is  1  in  ^— ^~ 

o.o 

=  1  in  10.4.  The  piezometric  line  DFC  is  drawn  up  from  D  parallel 
to  AB.  The  area  of  uplift  is  DGHEF;  that  part  of  the  uplift 
below  the  line  DE  is  however  accounted  for  by  assuming  all 
masonry  situated  below  El.  27.50  as  reduced  in  weight  by  flota- 
tion, leaving  the  area  DEF  as  representing  the  uplift  still  unac- 
counted for. 

Beyond  the  filter  is  a  21-meter  length  of  impervious  apron  con- 
sisting of  clay  puddle  covered  by  stone  paving,  which  abuts  on  a 
masonry  subsidiary  weir.  This  wall  holds  the  water  up  one  meter 
in  depth  and  so  reduces  the  head  to  that  extent,  with  the  further 
addition  of  the  depth  of  film  passing  over  the  crest  at  low  water 
which  is  .5  meter,  total  reduction  1.50  meters. 

This  is  the  first  instance  of  the  use  of  puddle  in  a  fore  apron, 
or  talus;  its  object  is,  by  the  introduction  of  an  impervious  rear 
apron  21  meters  long,  to  prevent  the  subsidiary  weir  wall  from  being 
undermined.  The  head  being  1J  meters,  the  length  of  travel 
required,  taking  c  as  18,  will  be  18X1.5  =  27  meters.  The  actual 
length  of  travel  provided  is  vertical  15,  horizontal  41,  total  56 
meters,  more  than  double  what  is  strictly  requisite.  The  long 
hearth  of  solid  masonry  which  is  located  below  the  subsidiary  drop 
wall  is  for  the  purpose  of  withstanding  scour  caused  by  the  over- 
fall. Beyond  this  is  the  talus  of  riprap  20  meters  wide  and  a  row 
of  sheet  piling.  The  total  length  of  the  floor  of  this  work  is  364 
feet,  with  three  rows  of  sheet  piles.  That  of  the  Assiut  barrage 
is  216  feet  with  two  rows  of  sheeting.  The  difference  in  head  is 
half  a  meter  only,  so  that  certain  unknown  conditions  of  flood  or 
that  of  the  material  in  the  bed  must  exist  to  account  for  the  excess. 

146.  American  vs.  Indian  Treatment.  In  American  regulat- 
ing works  it  is  generally  the  fashion  where  entire  closure  is  required 
to  locate  the  draw  gates  and  their  grooves  inside  the  panel  or  bulk- 
head wall  that  closes  the  upper  part  of  the  regulator  above  the 
sluice  openings.  Thus  when  the  gates  are  raised  they  are  concealed 
behind  the  panel  walls.  In  Indian  practice  the  gate  grooves  in  the 
piers  are  generally  located  outside  the  bulkhead  wall;  thus  when 
hoisted,  the  gates  are  visible  and  accessible.  Fig.  117  is  from  a 


DAMS  AND  WEIRS 


197 


photograph  of  a  branch  head,  illustrating  this.  The  work  is  of 
reinforced  concrete  as  can  be  told  from  the  thinness  of  the  piers. 
In  an  Indian  work  of  similar  character  the  pier  noses  would  project 
well  beyond  the  face  wall  of  the  regulator  and  the  gates  would  be 
raised  in  front,  not  behind  it. 

The  use  of  double  gates  is  universal  in  Eastern  irrigation  works; 
they  have  the  following  unquestionable  advantages  over  a  single 
gate:  First,  less  power  for  each  is  required  to  lift  two  gates  than 
one;  second,  when  hoisted  they  can  be  stacked  side  by  side  and  so 
the  pier  can  be  reduced  in  height;  third,  where  sand  or  silt  is  in 
suspension,  surface  water  can  be  tapped  by  leaving  the  lower  leaf 
down  while  the  upper  is  raised;  and  fourth,  regulation  is  made  easier. 


Fig.  117.     Typical  American  Regulating  Sluices  in  Reinforced- 
Concrete  Weir 

In  the  Khanki  weir  sluices,  Fig.  112,  3  catcs  running  in  3  grooves 
are  employed. 

147.  Length  of  Spans.  In  designing  open  dams  the  spans 
should  be  made  as  large  as  convenient,  the  tendency  in  modern 
design  is  to  increase  the  spans  to  30  feet  or  more;  the  Laguna  weir 
sluices  are  33|  feet  wide  and  the  Merala  31  feet.  The  thickness  of 
the  piers  is  a  matter  of  judgment  and  is  best  expressed  as  some 
function  of  the  span,  the  depth  of  water  by  which  the  height  of  the 
piers  is  regulated,  forms  another  factor. 

The  depth  of  water  upheld  regulates  the  thickness  more  than 
the  length.  The  length  should  be  so  adjusted  that  the  resultant 
line  of  pressure  combined  of  the  weight  of  one  pier  and  arch,  or 
superstructure  and  of  the  water  pressure  acting  on  one  span  falls 
within  the  middle  third  of  the  base. 


198  DAMS  AND  WEIRS 

For  example  take  the  Assiut  regulator,  Fig.  115.  The  con- 
tents of  one  pier  and  span  allowing  for  uplift  is  roughly  390  cubic 
meters  of  masonry,  an  equivalent  to  1000  tons.  The  incidence  of 
W  is  about  2  meters  from  the  middle  third  downstream  boundary. 

The  moment  of  the  weight  about  this  point  is  therefore  1000X 
2=2000  meter  tons.  Let  H  be  depth  of  water  upstream,  and  h 

downstream,  then  the  overturning  moment  is  expressed  by  - — 

o 

Here  H  =  5,h  =  2  meters,  w  =  l.l  tons  per  cubic  meter,  the  length  I 

f  .    n  (125-8)Xl.lX7 

or  one  span  is  7  meters;  then  the  moment  =—  —  =  150 

6 

meter  tons.  The  moment  of  resistance  is  therefore  immensely  in  excess 
of  the  moment  of  water  pressure.  The  height  of  the  pier  is  however 
governed  by  the  high  flood  level,  the  width  by  the  necessity  of  a 
highway  bridge.  At  full  flood  nearly  the  whole  of  the  pier  will  be 
immersed  in  water  and  so  lose  weight.  There  is  probably  some 
intermediate  stage  when  the  water  pressure  will  be  greater  than 
that  estimated,  as  would  be  the  case  if  the  gates  were  left  closed 
while  the  water  topped  them  by  several  feet,  the  water  downstream 
not  having  had  time  to  rise  to  correspond. 

148.  Moments  for  Hindia  Barrage.  In  the  case  of  the  Hindia 
barrage,  Fig.  116,  77  =  5  meters,  h  =  1.5,  then 

,f     (125-3.4)X1. 1X6.50 
M=—  —  =  145  meter  tons 

6 

The  weight  of  one  span  is  estimated  at  180  tons.  Its  moment 
about  the  toe  of  the  base  is  about  180X6.5  =  1170  meter  tons. 

The   factor   of    safety   against    overturning   is   therefore    :     —  =8. 

145 

The  long  base  of  these  piers  is  required  for  the  purpose  of 
distributing  the  load  over  as  wide  an  area  as  possible  in  order  to 
reduce  the  unit  pressure  to  about  one  long  ton  per  square  foot. 
This  is  also  partly  the  object  of  the  deep  mass  foundation.  The 
same  result  could  doubtless  be  attained  with  much  less  material  by 
adopting  a  thin  floor  say  two  or  three  feet  thick,  reinforced  by  steel 
rods  so  as  to  ensure  the  distribution  of  the  weight  of  the  super- 
structure evenly  over  the  whole  base.  It  seems  to  the  writer  that 
the  Assiut  barrage  with  its  mass  foundation  having  been  a  success 


DAMS  AND  WEIRS 


199 


200 


DAMS  AND  WEIRS 


TABLE  II 
Pier  Thickness — Suitable  for  Open  Partial   Regulators  and  Weir  Sluices 


DEPTHS  OF  WATER 

SPAN 

15  FEET 

20  FEET 

25  FEET 

30  FEET 

M. 

T. 

M. 

T. 

M. 

T. 

M. 

T. 

10  feet 

.25 

2.5 

.27 

2.7 

.29 

2.9 

.31 

3.1 

15  feet 

.24 

3.6 

.26 

3.9 

.28 

4.2 

.30 

4.5 

20  feet 

.23 

4.6 

.25 

5.0 

.27 

5.4 

.29 

5.8 

25  feet 

.21 

5.3 

.-24 

6.0 

.26 

6.5 

.28 

7.0 

M  is  multiplier  of  span  for  thickness  T. 


as  regards  stability,  is  no  reason  why  a  heavy  style  of  construction 
such  as  this  should  be  perpetuated. 

149..  North  Mon  Canal.  In  Fig.  118  is  shown  the  head  works 
of  the  Mon  right  canal  in  Burma.  The  weir  is  of  type  A,  with  crest 
shutters  and  sluices  of  large  span  controlled  by  draw  gates.  In  the 
canal  head,  the  gates  are  recessed  behind  the  face  wall  as  in  American 
practice. 

150.  Thickness  of  Piers.    Table  II,  though  purely  empirical 
will  form  a  useful  guide  of  thickness  of  piers  in  open  dams  or  partial 
regulators. 

If  reinforced,  very  considerable  reduction  can  be  made  in  the 
thickness  of  piers,  say  f ,  but  for  this  class  of  river  work  a  heavy 
structure  is  obligatory. 

151.  Advantages  of  Open  Dams.     Open  dams  have  the  follow- 
ing advantages  over  solid  weirs,  or  combinations  of  solid  arid  over- 
fall dams:     First,  the  river  bed  is  not  interfered  with  and  conse- 
quently the  heading  up  and  scour  is  only  that  due  to  the  obstruction 
of  the  piers,  which  is  inconsiderable.     This  points  to  the  value  of 
wide  spans.     Second,  the  "river  low"  supply  is  under  complete 
control.     Third,  a  highway  bridge  across  the  river  always  forms 
part  of  the  structure  which  in  most  countries  is  a  valuable  asset. 

Open  dams,  on  the  other  hand,  are  not  suitable  for  torrential 
rivers  as  the  Himalayan  rivers  near  their  points  of  debouchure 
from  the  mountains,  or  wherever .  such  detritus  as  trees,  logs,  etc., 
are  carried  down  in  flood  time. 


DAMS  AND  WEIRS  201 

152.  Upper  Coleroon  Regulator.  Fig.  119  is  from  a  photo- 
graph of  a  regulating  bridge  on  the  upper  Coleroon  River  in  the 
Madras  Presidency,  southern  India.  Originally  a  weir  of  type  A  was 
constructed  at  this  site  in  conjunction  with  a  bridge.  The  constric- 
tion of  the  discharge  due  to  the  drop  wall,  which  was  six  feet  high, 
and  the  piers  of  the  bridge,  caused  a  very  high  afflux  and  great  scour 
on  the  talus.  Eventually  the  drop  wall  was  cleared  away  altogether, 
the  bridge  piers  were  lengthened  upstream  and  fitted  with  grooves 
and  steel  towers,  and  counterweighted  draw  gates  some  7  feet  deep 


Fig.  119.     View  of  Regulating  Bridge  on  the  Upper  Coleroon  River,  Southern  India 

took  the  place  of  the  drop  wall.  In  the  flood  season  the  gates  can 
be  raised  up  to  the  level  of  the  bridge  parapet  quite  clear  of  the 
flood.  The  work  was  thus  changed  from  one  of  a  weir  of  type  A, 
to  an  open  dam.  The  original  weir  and  bridge  were  constructed 
about  half  a  century  ago. 

153.  St.  Andrew's  Rapids  Dam.  Another  class  of  semi-open 
dam  consists  of  a  permanent  low  floor  or  dwarf  weir  built  across 
the  river  bed  which  is  generally  of  rock,  and  the  temporary  dam- 
ming up  of  the  water  is  effected  by  movable  hinged  standards  being 
lowered  from  the  deck  of  an  overbridge,  which  standards  support 


202 


DAMS  AND  WEIRS 


DAMS  AND  WEIRS  203 

either  a  rolled  reticulated  curtain  let  down  to  cover  them  or  else  a 
steel  sliding  shutter  mounted  on  rollers. 

The  St.  Andrew's  Rapids  dam,  Fig.  120,  a  quite  recent  construc- 
tion, may  be  cited  as  an  example.  The  object  of  the  dam  is  to  raise 
the  water  in  the  Red  River,  Manitoba,  to  enable  steamboats  to  navi- 
gate the  river  from  Winnipeg  City  to  the  lake  of  that  name.  To 
effect  this  the  water  level  at  the  rapids  has  to  be  raised  20  feet 
above  L.  W.  L.  and  at  the  same  time,  on  account  of  the  accumu- 
lation of  ice  brought  down  by  the  river,  a  clear  passage  is  a  necessity. 
The  Red  River  rises  in  the  South,  in  the  State  of  North  Dakota 
where  the  thaw  sets  in  much  earlier  than  at  Lake  Winnipeg,  con- 
sequently freshets  bring  down  masses  of  ice  when  the  river  and  lake 
are  both  frozen. 

Camere  Type  of  Dam.  The  dam  is  of  the  type  known  as  the 
Camere  curtain  dam,  the  closure  being  effected  by  a  reticulated 
wooden  curtain,  which  is  rolled  up  and  down  the  vertical  frames 
thereby  opening  or  closing  the  vents.  It  is  a  French  invention, 
having  been  first  constructed  on  the  Seine.  The  principle  of  this 
movable  dam  consists  in  a  large  span  girder  bridge,  from  which 
vertical  hinged  supports  carrying  the  curtain  frames  are  let  drop 
on  to  a  low  weir.  When  not  required  for  use  these  vertical  girders 
are  hauled  up  into  a  horizontal  position  below  the  girder  bridge  and 
fastened  there.  In  fact,  the  principle  is  very  much  like  that  of  a 
needle  dam.  The  river  is  800  feet  wide,  and  the  bridge  is  of  six- 
spans  of  138  feet. 

The  bridge  is  composed  of  three  trusses,  two  of  which  are  free 
from  internal  cross-bracing,  and  carry  tram  lines  with  all  the  work- 
ing apparatus  of  several  sets  of  winches  and  hoists  for  manipulating 
the  vertical  girders  and  the  curtain;  the  third  truss  is  mainly  to 
strengthen  the  bridge  laterally,  and  to  carry  the  hinged  ends  of  the 
vertical  girders. 

It  will  be  understood  that  the  surface  exposed  to  wind  pressure 
is  exceptionally  great,  so  that  the  cross-bracing  is  absolutely  essen- 
tial, as  is  also  the  lateral  support  afforded  by  a  heavy  projection  of 
the  pier  itself  above  floor  level. 

In  the  cross-section  it  will  be  seen  that  there  is  a  footbridge 
opening  in  the  pier.  This  footbridge  will  carry  winches  for  wind- 
ing and  unwinding  the  curtains,  and  is  formed  by  projections  thrown 


204 


DAMS  AND  WEIRS 


out  at  the  rear  of  each  group  of  frames.  It  will  afford  through 
communication  by  a  tramway.  The  curtains  can  be  detached 
altogether  from  the  frames  and  housed  in  a  chamber  in  the  pier 
clear  of  the  floodline. 

The  lower  part  of  the  work  consists  of  a  submerged  weir  of 
solid  construction  which  runs  right  across  the  river;  its  crest  is  7 
feet  6  inches  above  L.  W.  L.  at  El.  689.50.  The  top  of  the  curtains 
to  which  water  is  upheld  is  El.  703.6,  or  14  feet  higher.  The  dam 
actually  holds  up  31  feet  of  water  above  the  bed  of  the  river. 


Half  rront  Elevation 
Fig.  121.     Lauchli  Automatic  Sluice  Gate 

This  system  is  open  to  the  following  objections:  First,  the 
immense  expense  involved  in  a  triple  row  of  steel  girders  of  large 
span  carrying  the  curtains  and  their  apparatus;  and  second,  the 
large  surface  exposure  to  wind  which  must  always  be  a  menace  to 
the  safety  of  the  curtains. 

It  is  believed  that  the  raising  of  the  water  level  could  be  effected 
for  a  quarter  of  the  cost  if  not  much  less,  by  adopting  a  combination 
of  the  system  used  in  the  Folsam  weir,  Fig.  50,  with  that  in  the 
Dhukwa  weir,  Fig.  52,  viz,  hinged  collapsible  gates  which  could  be 
pushed  up  or  lowered  by  hydraulic  jacks  as  required.  The  existing 
lower  part  of  the  dam  could  be  utilized  and  »  subway  constructed 


DAMS  AND  WEIRS  205 

through  it  for  cross  communication  and  accommodation  for  the 
pressure  pipes,  as  is  the  case  in  the  Dhukwa  weir.  This  arrange- 
ment which  is  quite  feasible  would,  it  is  deemed,  be  an  improvement 
on  the  expensive,  complicated,  and  slow,  Camere  curtain  system. 

1 54.  Automatic  Dam  or  Regulator.  Mr.  Lauchli  of  New  York, 
writing  for  Engineering  News,  describes  a  new  design  for  automatic 
regulators,  as  follows: 

In  Europe  there  has  been  in  operation  for  some  time  a  type  of  automatic 
dam  or  sluice  gate  which  on  account  of  its  simplicity  of  construction,  adapt- 


Fig.  122.     View  of  Lauchli  Automatic  Dam  Which  Has  Been  for  Several  Years 
in  Successful  Operation  in  Europe 

ability  to  existing  structures,  exact  mathematical  treatment,  and  especially 
its  successful  operation,  deserves  to  attract  the  attention  of  the  hydraulic  engineer 
connected  with  the  design  of  hydroelectric  plants  or  irrigation  works.  Fig. 
121  shows  a  cross-section  and  front  elevation  of  one  of  the  above-mentioned 
dams  now  in  course  of  construction,  and  the  view  in  Fig.  122  gives  an  idea  of 
a  small  automatic  dam  of  the  same  type  which  has  been  in  successful  operation 
for  several  seasons,  including  a  severe  winter,  and  during  high  spring  floods. 

Briefly  stated,  the  automatic  dam  is  composed  of  a  movable  part  or 
panel,  resting  at  the  bottom  on  a  knife  edge,  and  fastened  at  the  top  to  a  com- 
pensating roller  made  of  steel  plate  and  filled  with  concrete.  This  roller  moves 
along  a  track  located  at  each  of  its  ends,  and  is  so  designed  as  to  take,  at  any 
height  of  water  upstream,  a  position  such  as  will  give  the  apron  the  inclination 
necessary  for  discharging  a  known  amount  of  water,  and  in  so  doing  will  keep 
the  upper  pool  at  a  constant  fixed  elevation. 

With  the  roller  at  its  highest  position  the  panel  lies  horizontally,  and 
the  full  section  is  then  available  for  discharging  water.  Any  debris,  such  as 


206  DAMS  AND  WEIRS 

trees,  or  ice  cakes,  etc.,  will  pass  over  the  dam  without  any  difficulty,  even 
during  excessive  floods,  as  the  compensating  roller  is  located  high  above  extreme 
flood  level. 

The  dam  now  in  course  of  construction  is  located  on  the  river  Grafenauer 
Ohe,  in  Bavaria,  and  will  regulate  the  water  level  at  the  intake  of  a  paper  mill, 
located  at  some  distance  from  the  power  house.  The  dam  has  a  panel  24.27 
ft.  long,  6.85  ft.  high,  and  during  normal  water  level  will  discharge  1400  cu.  ft. 
per  sec.,  while  at  flood  time  it  will  pass  3,530  cu.  ft.  per  sec.  of  water.  As  shown 
in  Fig.  121,  the  main  body  of  the  dam  is  made  of  a  wooden  plank  construction 
laid  on  a  steel  frame.  The  panel  is  connected  with  the  compensating  roller 
at  each  end  by  a  flexible  steel  cable  wound  around  the  roller  end,  and  then 
fastened  at  the  upper  part  of  the  roller  track  to  an  eyebolt.  A  simple  form  of 
roof  construction  protects  the  roller  track  from  rain  and  snow.  The  panel 
is  made  watertight  at  each  extremity  by  means  of  galvanized  sheet  iron  held 
light  against  the  abutments  by  water  pressure.  This  type  of  construction  has 
so  far  proved  to  be  very  effective  as  to  watertightness. 

It  may  be  needless  to  point  out  that  this  type  of  dam  can  also  be  fitted 
to  the  crest  of  overflow  dam  of  ordinary  cross-section,  and  then  fulfill  the  duty 
of  movable  flashboards. 

The  probability  is  that  this  type  will  become  largely  used  in 
the  future.  A  suggested  improvement  would  be  to  abolish  the  cross 
roller  having  instead  separate  rollers  on  each  pier  or  abutment, 
working  independently.  There  will  then  be  no  practical  limit  to 
the  span  adopted. 


INDEX 


Aprons . 

decrease  uplift,  rear  aprons 71 

fore,  base  of  dam  and 98 

hearth  and  anchored 150 

porous  fore 173,  175 

rear ..  159,  161 

riprap  to  protect 164 

sloping 169 

uplift,  affect 70 

Arched  dams 101 

characteristics 101 

crest  width 104 

examples 104 

Barossa 111 

Bear  Valley .__  104 

Burrin  Juick  subsidiary 112 

Lithgow 112 

Pathfinder  __                                                                                                ._.  104 

Shoshone .__  107 

Sweetwater ___  109 

profiles __  103 

correct .  _  _  103 

theoretical  and  practical - .  _  102 

variable  radii,  with .__  112 

vertical  water  loads,  support  of _  _  _  104 

Arrow  Rock  dam __  67 

Assiut  barrage 192,  194 

Automatic  dam  or  regulator 205 

B 

Barossa  dam 111 

Barrages .__  182 

Bassano  dam 146 

Bear  Valley  dam 104 

Burrin  Juick  subsidiary  dam 112 

C 

Castlewood  weir 96 

D 

Damietta  and  Rosetta  weirs 179 

Dams  and  weirs  __  1 


INDEX 

PAGE 

Dams  and  weirs — continued 

arched 101 

definition 1 

gravity  _  _ . . 2 

gravity  overfall „__  75 

hollow  slab  buttress 136 

multiple  arch  or  hollow  arch  buttress. __   113 

open  dams  or  barrages 182 

submerged  weirs  founded  on  sand 151 

Dehri  weir 178 

Dhukwa  weir 90 

E 

Ellsworth  dam 136 

F 

Folsam  weir 85 

G 

Granite  Reef  weir 92 

Gravity  dams 2 

design 4 

analytical  method 18,  34,  43 

broken  line  profiles,  treatment  for 41 

calculation,  method  of 11 

crest,  high  and  wide 13 

crest  width 9 

failure  by  sliding  or  shear,  security  against 31 

graphical  method 16 

Haessler's  method 1 36,  42 

height,  variation  of -  13 

influence  lines 31 

maximum  stress,  formulas  for 27 

elementary  profile,  application  to 28 

limiting  height  by 29 

pressure  area  in  inclined  back  dam,  modified  equivalent. .  _  .  37 

pressure  distribution 23,  25,  26 

pressure  limit,  maximum 27 

pressures  in  figures,  actual 34 

profile,  practical 8 

profile,  theoretical 4 

profiles,  curved  back 39 

rear  widening 10 

shear  and  tension,  internal 30 

stepped  polygon 37 

vertical  component —  22 

discussion 2 

graphical  calculations  _  _ . 

"middle  third"  and  limiting  stress 

pressure  of  water  on  wall 2 


INDEX 

PAGE 

Gravity  dams — continued 
discussion 

stress  limit,  compressive 4 

examples 53 

Arrow  Rock 67 

Assuan 59 

Burrin  Juick 65 

Cheescman  Lake 53 

Cross  River  and  Ashokan 65 

New  Croton 58 

Roosevelt 56 

foundat  ions,  special 69 

aprons  affect  uplift 70 

aprons  decrease  uplift,  rear -  _  _  71 

ice  pressure,  gravity  dam  reinforced  against .  _  _  73 

rock  below  gravel .. 72 

"high" --.  43 

base  of  dam,  silt  against 50 

partial  overfall 52 

pentagonal  profile  to  be  widened 47 

pressure,  ice .  -  _  51 

toe  of  dam,  filling  against 50 

Gravity  overfall  dams  or  weirs 75 

American  dams  on  pervious  foundations .__  97 

analytical  method 88 

base  width,  approximate 77 

characteristics 75 

crest  width,  approximate '- 77 

depth  of  overfall,  calculation  of 82 

examples .  _  _  83 

Castlewood _  _  _  96 

Dhukwa__                                                                                               ._.  90 

Folsam 1 85 

Granite  Reef 92 

Mariquina 92 

Nira 95 

"Ogee" 85 

St.  Maurice  River _  _  _  99 

fore  apron,  base  of  dam  and 98 

graphical  process 78 

hydraulic  conditions 93 

pressure,  moments  of  _ .                                                                      81 

water  level,  pressures  affected  by 79 

Guayabal  dam 141 

H 

Haessler's  method  _.  __36,  42 

Hindia  barrage.  _                                                                                              _ _  194,  198 

Hollow  slab  buttress  dam__                                                                                    _  136 


INDEX 

PAGE 

Hollow  slab  buttress  dam — continued 

description ' 136 

examples 146 

Bassano 146 

Ellsworth !.___, 136 

Guayabal 141 

fore  slope,  steel  in 139 

foundation  foredeck,  pressure  on 149 

baffles 150 

buttresses 150 

hearth  or  anchored  apron_  _    : 150 

reinforced  concrete,  formulas  for 137 

slab  deck  compared  with  arch  deck 140 


K 
Khanki  weir  _ .  _  171 


L 

Laguna  weir 179 

Lithgow  dam 112 


M 

Mariquina  weir 92 

Merala  weir 171 

Mir  Alam  dam 114 

Multiple  arch  or  hollow  arch  buttress  dams 113 

arch,  crest  width  of 125 

arches,  differential 126 

design , 122 

examples 

Belubula 118 

Big  Bear  Valley 131 

Mir  Alam 114 

Ogden 120 

inclination  of  arch  to  vertical : 119 

pressure 

flood 129 

foundation,  on 125 

water,  reverse 124 

stresses 1 117 

value  ..  -  113 


N 

Narora  weir 167 

Niraweir__  95 


INDEX 


"Ogee"  gravity  overfall  dam 85 

Okhla  and  Madaya  weirs 177 

Open  dams  or  barrages 182 

advantages 200 

American  vs.  Indian  treatment 196 

automatic 205 

Corbett  dam,  weir  sluices  of 189 

definition 182 

examples 

Assiut 192,  194 

Hindia 194,  198 

North  Mon 200 

Upper  Coleroon 201 

St.  Andrew's  Rapids 201 

CarneYe"  type 203 

heavy  construction 190 

moments  for  Hindia  barrage 198 

piers,  thickness  of 200 

regulator,  Upper  Coleroon 201 

river  regulators,  features  of 193 

rivers,  across 192 

spans,  length  of 197 

weir  scouring  sluice,  example  of 185 

weir  scouring  sluice  on  sand _ 190 

P 

Pathfinder  dam 104 

S 

St.  Andrew's  Rapids  dam 201 

Shoshone  dam 107 

Stepped  polygon 37 

Submerged  weirs  on  sand 151 

apron,  rear 159,  161 

computations,  simplifying 156 

crest  shutters 182 

description 151 

examples 

Damietta  and  Rosetta 179 

Dehri 178 

Khanki 171 

Laguna 179 

Merala .__.  171 

Narora 167 

Okhla  and  Madaya 177 

fore  aprons,  porous 173 

core  walls,  divided  by 175 


INDEX 

PAGE 

Submerged  weirs  on  sand — continued 

hydraulic  flow,  laws  of 152 

percolation 

coefficient  of 153 

beneath  dam_ 152 

values  of  coefficient  of 155 

vertical  obstruction  to 159 

pervious  dam,  paradox  of  a 181 

riprap 164 

safety,  criterion  for 154 

sloping  apron 169 

stability,  governing  factor  for 153 

Sweetwater  dam 109 

T 

Tables 

pier  thickness 200 

values  of  L{  or  talus  width '.__                        163 

U 

Upper  Coleroon  regulator 201 


THIS 


OVERDUE- 


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FEB 
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wFQ 


LD21 


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